The diagram is given below:
Let A B C ABC A BC be the triangle. Let O O O be the circumcenter and I , I, I , the incenter.
Clearly, O A = O B = O C = R , I E = r [ I E ⊥ A B ] OA = OB = OC = R, IE=r[IE\perp AB] O A = OB = OC = R , I E = r [ I E ⊥ A B ]
Let O M ⊥ B C OM\perp BC OM ⊥ BC then ∠ B O M = ∠ C O M = A \angle BOM = \angle COM = A ∠ BOM = ∠ COM = A
Now, O A = R , A I = r cosec A 2 = 4 R sin A 2 sin B 2 sin C 2 sin A 2 OA = R, AI = r\cosec\frac{A}{2} = \frac{4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\sin\frac{A}{2}} O A = R , A I = r cosec 2 A = s i n 2 A 4 R s i n 2 A s i n 2 B s i n 2 C
= 4 R sin B 2 sin C 2 = 4R\sin\frac{B}{2}\sin\frac{C}{2} = 4 R sin 2 B sin 2 C
∠ O A B = ∠ O B A = B − ( 9 0 ∘ − A ) = A + B − 9 0 ∘ = 9 0 ∘ − C \angle OAB = \angle OBA = B - (90^\circ - A) = A + B - 90^\circ = 90^\circ - C ∠ O A B = ∠ OB A = B − ( 9 0 ∘ − A ) = A + B − 9 0 ∘ = 9 0 ∘ − C
∴ ∠ O A I = ∠ B A I − ∠ B A O = A 2 − ( 9 0 ∘ − C ) \therefore \angle OAI = \angle BAI - \angle BAO = \frac{A}{2} - (90^\circ - C) ∴ ∠ O A I = ∠ B A I − ∠ B A O = 2 A − ( 9 0 ∘ − C )
= C − B 2 = \frac{C - B}{2} = 2 C − B
Applying cosine law in △ O A I , \triangle OAI, △ O A I ,
cos C − B 2 = O A 2 + A I 2 − O I 2 2 O A . A I \cos\frac{C - B}{2} = \frac{OA^2 + AI^2 - OI^2}{2OA.AI} cos 2 C − B = 2 O A . A I O A 2 + A I 2 − O I 2
= R 2 + 16 R 2 sin 2 B 2 sin 2 C 2 − O I 2 2. R . 4 R sin B 2 sin C 2 = \frac{R^2 + 16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} - OI^2}{2.R.4R\sin\frac{B}{2}\sin\frac{C}{2}} = 2. R .4 R s i n 2 B s i n 2 C R 2 + 16 R 2 s i n 2 2 B s i n 2 2 C − O I 2
O I 2 = R 2 [ 1 + 8 sin B 2 sin C 2 { 2 sin B 2 sin C 2 − cos ( B − C 2 ) } ] OI^2 = R^2\left[1 + 8\sin\frac{B}{2}\sin\frac{C}{2}\left\{2\sin\frac{B}{2}\sin\frac{C}{2} - \cos\left(\frac{B -
C}{2}\right)\right\}\right] O I 2 = R 2 [ 1 + 8 sin 2 B sin 2 C { 2 sin 2 B sin 2 C − cos ( 2 B − C ) } ]
= R 2 [ 1 + 8 sin B 2 sin C 2 { cos B − C 2 − cos B + C 2 − cos B − C 2 } ] = R^2\left[1 + 8\sin\frac{B}{2}\sin\frac{C}{2}\left\{\cos\frac{B - C}{2} - \cos\frac{B + C}{2} - \cos\frac{B -
C}{2}\right\}\right] = R 2 [ 1 + 8 sin 2 B sin 2 C { cos 2 B − C − cos 2 B + C − cos 2 B − C } ]
= R 2 [ 1 − 8 sin A 2 sin B 2 sin C 2 ] = R^2\left[1 - 8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right] = R 2 [ 1 − 8 sin 2 A sin 2 B sin 2 C ]
= R 2 − 2 R r = R^2 - 2Rr = R 2 − 2 R r
Let us prove the necessary condition for the second part.
Let b b b be the A.M. of a a a and c c c i.e. 2 b = a + c 2b = a + c 2 b = a + c
⇒ 2 sin B = sin A + sin C ⇒ 2.2 sin B 2 cos B 2 = 2 sin A + C 2 cos A − C 2 \Rightarrow 2\sin B = \sin A + \sin C \Rightarrow 2.2\sin\frac{B}{2}\cos\frac{B}{2} = 2\sin\frac{A + C}{2}\cos\frac{A -
C}{2} ⇒ 2 sin B = sin A + sin C ⇒ 2.2 sin 2 B cos 2 B = 2 sin 2 A + C cos 2 A − C
⇒ 2 sin B 2 = cos A − C 2 ⇒ 2 cos A + C 2 = cos A − C 2 \Rightarrow 2\sin\frac{B}{2} = \cos\frac{A - C}{2} \Rightarrow 2\cos\frac{A + C}{2} = \cos\frac{A - C}{2} ⇒ 2 sin 2 B = cos 2 A − C ⇒ 2 cos 2 A + C = cos 2 A − C
cos B I O = B I 2 + I O 2 − B O 2 2 B I . I O \cos BIO = \frac{BI^2 + IO^2 - BO^2}{2BI.IO} cos B I O = 2 B I . I O B I 2 + I O 2 − B O 2
Now B I 2 + I O 2 − B O 2 = r 2 cosec 2 B 2 + R 2 − 2 r R − R 2 BI^2 + IO^2 - BO^2 = r^2\cosec^2\frac{B}{2} + R^2 - 2rR - R^2 B I 2 + I O 2 − B O 2 = r 2 cosec 2 2 B + R 2 − 2 r R − R 2
= r 2 cosec 2 B 2 − 2 r R = r^2\cosec^2\frac{B}{2} - 2rR = r 2 cosec 2 2 B − 2 r R
= r . 4 R sin A 2 sin B 2 sin C 2 sin 2 B 2 − 2 r R = \frac{r.4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\sin^2\frac{B}{2}} - 2rR = s i n 2 2 B r .4 R s i n 2 A s i n 2 B s i n 2 C − 2 r R
= 2 r R . 2 sin A 2 sin C 2 sin B 2 − 2 r R = \frac{2rR.2\sin\frac{A}{2}\sin\frac{C}{2}}{\sin\frac{B}{2}} - 2rR = s i n 2 B 2 r R .2 s i n 2 A s i n 2 C − 2 r R
= 2 r R [ cos A − C 2 − cos A + C 2 ] sin B 2 − 2 r R = \frac{2rR\left[\cos\frac{A - C}{2} - \cos\frac{A + C}{2}\right]}{\sin\frac{B}{2}} - 2rR = s i n 2 B 2 r R [ c o s 2 A − C − c o s 2 A + C ] − 2 r R
= 2 r R 2 cos A + C 2 − cos A + C 2 sin B 2 − 2 r R = 0 = 2rR\frac{2\cos\frac{A + C}{2}- \cos\frac{A + C}{2}}{\sin\frac{B}{2}} - 2rR = 0 = 2 r R s i n 2 B 2 c o s 2 A + C − c o s 2 A + C − 2 r R = 0
Thus, △ B I O \triangle BIO △ B I O is a right angled triangle.
Now the sufficient condition can be proved similarly.
A 2 + B 2 = π 2 − C 2 \frac{A}{2} + \frac{B}{2} = \frac{\pi}{2} - \frac{C}{2} 2 A + 2 B = 2 π − 2 C
⇒ cot ( A 2 + B 2 ) = cot ( π 2 − C 2 ) \Rightarrow \cot\left(\frac{A}{2} + \frac{B}{2}\right) = \cot\left(\frac{\pi}{2} - \frac{C}{2}\right) ⇒ cot ( 2 A + 2 B ) = cot ( 2 π − 2 C )
⇒ cot A 2 cot B 2 − 1 cot A 2 + cot B 2 = tan C 2 = 1 cot C 2 \Rightarrow \frac{\cot\frac{A}{2}\cot\frac{B}{2} - 1}{\cot\frac{A}{2} + \cot\frac{B}{2}} = \tan\frac{C}{2} =
\frac{1}{\cot\frac{C}{2}} ⇒ c o t 2 A + c o t 2 B c o t 2 A c o t 2 B − 1 = tan 2 C = c o t 2 C 1
⇒ cot A 2 + cot B 2 + cot C 2 = cot A 2 cot B 2 cot C 2 \Rightarrow \cot\frac{A}{2} + \cot \frac{B}{2} + \cot\frac{C}{2} = \cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2} ⇒ cot 2 A + cot 2 B + cot 2 C = cot 2 A cot 2 B cot 2 C
The diagram is given below:
H L = r 2 , I D = r ∴ I L = I D − L D = I D − H K = r − r 2 HL = r_2, ID=r \therefore IL = ID - LD = ID - HK = r - r_2 H L = r 2 , I D = r ∴ I L = I D − L D = I D − HK = r − r 2
In △ I H L , \triangle IHL, △ I H L ,
cot B 2 = H L I L = r 2 r − r 2 \cot\frac{B}{2} = \frac{HL}{IL} = \frac{r_2}{r - r_2} cot 2 B = I L H L = r − r 2 r 2
Similarly, cot A 2 = r 1 r − r 1 \cot\frac{A}{2} = \frac{r_1}{r - r_1} cot 2 A = r − r 1 r 1 and
cot C 2 = r 3 r − r 3 \cot\frac{C}{2} = \frac{r_3}{r - r_3} cot 2 C = r − r 3 r 3
Now following the result of previous problem
r 1 r − r 1 + r 2 r − r 2 + r 3 r − r 3 = r 1 r 2 r 3 ( r − r 1 ) ( r − r 2 ) ( r − r 3 ) \frac{r_1}{r - r_1} + \frac{r_2}{r - r_2} + \frac{r_3}{r - r_3} = \frac{r_1r_2r_3}{(r - r_1)(r - r_2)(r - r_3)} r − r 1 r 1 + r − r 2 r 2 + r − r 3 r 3 = ( r − r 1 ) ( r − r 2 ) ( r − r 3 ) r 1 r 2 r 3
The diagram is given below:
Let O O O be the circumcenter and P P P the orthocenter of the △ A B C . \triangle ABC. △ A BC . From geometry,
∠ B O F = ∠ C O F = A \angle BOF = \angle COF = A ∠ BOF = ∠ COF = A and A P = 2 O F = 2 R cos A , B F = R sin A AP = 2OF = 2R\cos A, BF = R\sin A A P = 2 OF = 2 R cos A , BF = R sin A
From right angled △ A D B \triangle ADB △ A D B
B D = c cos B , A D = c sin B BD = c\cos B, AD = c\sin B B D = c cos B , A D = c sin B
Now, P M = A D − ( A P + M D ) = c sin B − ( 2 R cos A + R cos A ) = c sin B − 3 R cos A PM = AD - (AP + MD) = c\sin B - (2R\cos A + R\cos A) = c\sin B - 3R\cos A PM = A D − ( A P + M D ) = c sin B − ( 2 R cos A + R cos A ) = c sin B − 3 R cos A
O M = F D = B F − B D = R sin A − c cos B OM = FD = BF - BD = R\sin A - c\cos B OM = F D = BF − B D = R sin A − c cos B
∴ tan θ = P M O M = c sin B − 3 R cos A R sin A − c cos B \therefore \tan\theta = \frac{PM}{OM} = \frac{c\sin B - 3R\cos A}{R\sin A - c\cos B} ∴ tan θ = OM PM = R s i n A − c c o s B c s i n B − 3 R c o s A
= 2 R sin C sin B − 3 R cos A R sin A − 2 R sin C cos B = \frac{2R\sin C\sin B - 3R\cos A}{R\sin A - 2R\sin C\cos B} = R s i n A − 2 R s i n C c o s B 2 R s i n C s i n B − 3 R c o s A
= 2 sin B sin C − 2 cos [ π − ( B + C ) ] sin [ π − ( B + C ) − 2 sin C cos B ] = \frac{2\sin B\sin C - 2\cos[\pi - (B + C)]}{\sin[\pi - (B + C)- 2\sin C\cos B]} = s i n [ π − ( B + C ) − 2 s i n C c o s B ] 2 s i n B s i n C − 2 c o s [ π − ( B + C )]
= 2 sin B sin C + 3 cos ( B + C ) sin ( B + C ) − 2 sin C cos B = \frac{2\sin B\sin C + 3\cos(B + C)}{\sin(B + C) - 2\sin C\cos B} = s i n ( B + C ) − 2 s i n C c o s B 2 s i n B s i n C + 3 c o s ( B + C )
= 3 cos B cos C − sin B sin C cos C sin B − sin C cos B = \frac{3\cos B\cos C - \sin B\sin C}{\cos C\sin B - \sin C\cos B} = c o s C s i n B − s i n C c o s B 3 c o s B c o s C − s i n B s i n C
= 3 − tan B tan C tan B − tan C = \frac{3 - \tan B\tan C}{\tan B - \tan C} = t a n B − t a n C 3 − t a n B t a n C
The diagram is given below:
Let A B C ABC A BC be the triangle. Let O O O and I I I be the circumcenter and in-center of the △ A B C . \triangle ABC. △ A BC .
Let P P P be the center and x , x, x , the radius of the circle drawn which touches the inscribed and circumscribed
circle, of △ A B C \triangle ABC △ A BC and the side B C BC BC externally. Let us join O P OP OP and extend up to Q . Q. Q . Let
I D ⊥ B C . ID\perp BC. I D ⊥ BC . Clearly, P P P will lie on the extended part of I D . ID. I D . Draw the line O N ON ON parallel to the
line I P . IP. I P . Join N P . NP. NP .
Clearly, O B = O C O Q = R , O P = O Q − P Q = R − x OB = OC OQ = R, OP = OQ - PQ = R - x OB = OCOQ = R , OP = OQ − PQ = R − x
O N = O M + M N = R cos A + x ON = OM + MN = R\cos A + x ON = OM + MN = R cos A + x
N P = M D = O M − C D = a 2 − r cot C 2 NP = MD = OM - CD = \frac{a}{2} - r\cot \frac{C}{2} NP = M D = OM − C D = 2 a − r cot 2 C
= a 2 − Δ s . s ( s − c ) Δ = a 2 − ( s − c ) = \frac{a}{2} - \frac{\Delta}{s}.\frac{s(s - c)}{\Delta} = \frac{a}{2} - (s - c) = 2 a − s Δ . Δ s ( s − c ) = 2 a − ( s − c )
= c − b 2 = \frac{c - b}{2} = 2 c − b
From right angled △ O N P , O P 2 = O N 2 + N P 2 \triangle ONP, OP^2 = ON^2 + NP^2 △ ONP , O P 2 = O N 2 + N P 2
( R − x ) 2 = ( R cos A + x ) 2 + ( c − b ) 2 (R - x)^2 = (R\cos A + x)^2 + \left(c - b\right)^2 ( R − x ) 2 = ( R cos A + x ) 2 + ( c − b ) 2
R 2 + x 2 − 2 R x = R 2 cos 2 A + x 2 + 2 R x cos A + ( c − b 2 ) 2 R^2 + x^2 - 2Rx = R^2\cos^2A + x^2 + 2Rx\cos A + \left(\frac{c - b}{2}\right)^2 R 2 + x 2 − 2 R x = R 2 cos 2 A + x 2 + 2 R x cos A + ( 2 c − b ) 2
2 R x ( 1 + cos A ) = R 2 ( 1 − cos 2 A ) − ( c − b 2 ) 2 2Rx(1 + \cos A) = R^2(1 - \cos^2A) - \left(\frac{c - b}{2}\right)^2 2 R x ( 1 + cos A ) = R 2 ( 1 − cos 2 A ) − ( 2 c − b ) 2
4 R x cos 2 A 2 = R 2 sin 2 A − ( b − c 2 ) 2 = a 2 4 − ( b − c ) 2 4 4Rx\cos^2\frac{A}{2} = R^2\sin^2A - \left(\frac{b - c}{2}\right)^2 = \frac{a^2}{4} - \frac{(b - c)^2}{4} 4 R x cos 2 2 A = R 2 sin 2 A − ( 2 b − c ) 2 = 4 a 2 − 4 ( b − c ) 2
4 R x s ( s − a ) b c = a + b − c 2 . a − b + c 2 = ( s − b ) ( s − c ) 4Rx\frac{s(s - a)}{bc} = \frac{a +b - c}{2}.\frac{a - b + c}{2} = (s - b)(s - c) 4 R x b c s ( s − a ) = 2 a + b − c . 2 a − b + c = ( s − b ) ( s − c )
x = Δ a tan 2 A 2 x = \frac{\Delta}{a}\tan^2\frac{A}{2} x = a Δ tan 2 2 A
The diagram is given below:
Since angles n the same segment of a circle are equal. ∴ ∠ B E D = ∠ B A D = A 2 \therefore \angle BED= \angle BAD = \frac{A}{2} ∴ ∠ BE D = ∠ B A D = 2 A
and ∠ B E F = ∠ B C F = C 2 \angle BEF = \angle BCF = \frac{C}{2} ∠ BEF = ∠ BCF = 2 C
Now ∠ D E F = ∠ B E F + ∠ B E D = C 2 + A 2 = C + A 2 = 9 0 ∘ − B 2 \angle DEF = \angle BEF + \angle BED = \frac{C}{2} + \frac{A}{2} = \frac{C + A}{2} = 90^\circ - \frac{B}{2} ∠ D EF = ∠ BEF + ∠ BE D = 2 C + 2 A = 2 C + A = 9 0 ∘ − 2 B
Similarly, D F E = 9 0 ∘ − C 2 , ∠ E D F = 9 0 ∘ − A 2 DFE = 90^\circ - \frac{C}{2}, \angle EDF = 90^\circ - \frac{A}{2} D FE = 9 0 ∘ − 2 C , ∠ E D F = 9 0 ∘ − 2 A
Now area of △ D E F = 1 2 D E . D F . sin ∠ E D F \triangle DEF = \frac{1}{2}DE.DF.\sin \angle EDF △ D EF = 2 1 D E . D F . sin ∠ E D F
Let R R R be the circum-radius of △ A B C \triangle ABC △ A BC then clearly R R R is also the circum-radius of △ D E F . \triangle DEF. △ D EF .
Applying sine rule in △ D E F , \triangle DEF, △ D EF , we have
D E sin D F E = D F sin D E F = E F sin E D F = 2 R \frac{DE}{\sin DFE} = \frac{DF}{\sin DEF} = \frac{EF}{\sin EDF} = 2R s i n D FE D E = s i n D EF D F = s i n E D F EF = 2 R
⇒ D E = 2 R sin D F E = 2 R sin ( 9 0 ∘ − C 2 ) = 2 R cos C 2 \Rightarrow DE = 2R\sin DFE = 2R\sin\left(90^\circ - \frac{C}{2}\right) = 2R\cos\frac{C}{2} ⇒ D E = 2 R sin D FE = 2 R sin ( 9 0 ∘ − 2 C ) = 2 R cos 2 C
Similarly, D F = 2 R cos B 2 DF = 2R\cos\frac{B}{2} D F = 2 R cos 2 B
So, area of △ D E F = 1 2 . 4 R 2 cos B 2 cos C 2 . sin ( 9 0 ∘ − A 2 ) \triangle DEF = \frac{1}{2}.4R^2\cos\frac{B}{2}\cos\frac{C}{2}.\sin\left(90^\circ - \frac{A}{2}\right) △ D EF = 2 1 .4 R 2 cos 2 B cos 2 C . sin ( 9 0 ∘ − 2 A )
= 2 R 2 cos A 2 cos B 2 cos C 2 = 2R^2\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} = 2 R 2 cos 2 A cos 2 B cos 2 C
= 2 R 2 s ( s − a ) b c s ( s − b ) c a s ( s − c ) a b = 2R^2\sqrt{\frac{s(s - a)}{bc}}\sqrt{\frac{s(s - b)}{ca}}\sqrt{\frac{s(s - c)}{ab}} = 2 R 2 b c s ( s − a ) c a s ( s − b ) ab s ( s − c )
= 2 R 2 s s ( s − a ) ( s − b ) ( s − c ) a b c = 2 R 2 s Δ a b c = R 2 . s . 4 Δ 2 a b c = R 2 s 2 a b c 4 Δ = \frac{2R^2s\sqrt{s(s - a)(s - b)(s - c)}}{abc} = \frac{2R^2s\Delta}{abc} = \frac{R^2.s.4\Delta}{2abc} = \frac{R^2s}{2\frac{abc}{4\Delta}} = ab c 2 R 2 s s ( s − a ) ( s − b ) ( s − c ) = ab c 2 R 2 s Δ = 2 ab c R 2 . s .4Δ = 2 4Δ ab c R 2 s
Here, are of △ A B C = Δ \triangle ABC = \Delta △ A BC = Δ
= R 2 s 2 R = R 2 s = \frac{R^2s}{2R} = \frac{R}{2}s = 2 R R 2 s = 2 R s
⇒ Δ D E F Δ A B C = R s 2 Δ = R 2 r \Rightarrow \frac{\Delta DEF}{\Delta ABC} = \frac{Rs}{2\Delta} = \frac{R}{2r} ⇒ Δ A BC Δ D EF = 2Δ R s = 2 r R
The diagram is given below:
Δ A ′ B ′ C ′ = Δ A B C − ( Δ A C ′ B ′ + Δ B A ′ C ′ + Δ C A ′ B ′ ) \Delta A'B'C' = \Delta ABC - (\Delta AC'B' + \Delta BA'C' + \Delta CA'B') Δ A ′ B ′ C ′ = Δ A BC − ( Δ A C ′ B ′ + Δ B A ′ C ′ + Δ C A ′ B ′ )
Δ A C ′ B ′ = 1 2 A C ′ . A B ′ sin A \Delta AC'B' = \frac{1}{2}AC'.AB'\sin A Δ A C ′ B ′ = 2 1 A C ′ . A B ′ sin A
∵ C C ′ \because CC' ∵ C C ′ is the internal bisector of ∠ C \angle C ∠ C
A C ′ C ′ B = A C C B = b a \frac{AC'}{C'B} = \frac{AC}{CB} = \frac{b}{a} C ′ B A C ′ = CB A C = a b
⇒ A C ′ = b k , C ′ B = a k , \Rightarrow AC' = bk, C'B = ak, ⇒ A C ′ = bk , C ′ B = ak , where k k k is some constant dependent on angles.
A C ′ + C ′ B = A B ⇒ c = a k + b k ⇒ k = c a + b AC' + C'B = AB \Rightarrow c = ak + bk \Rightarrow k = \frac{c}{a + b} A C ′ + C ′ B = A B ⇒ c = ak + bk ⇒ k = a + b c
∴ A C ′ = b c a + b \therefore AC' = \frac{bc}{a + b} ∴ A C ′ = a + b b c
Similarly, A B ′ = b c a + c AB' = \frac{bc}{a + c} A B ′ = a + c b c
Δ A C ′ B ′ = 1 2 b c a + b b c a + c sin A \Delta AC'B' = \frac{1}{2}\frac{bc}{a + b}\frac{bc}{a + c}\sin A Δ A C ′ B ′ = 2 1 a + b b c a + c b c sin A
Let Δ \Delta Δ be the area of △ A B C , \triangle ABC, △ A BC , then Δ = 1 2 b c sin A \Delta = \frac{1}{2}bc\sin A Δ = 2 1 b c sin A
∴ Δ A C ′ B ′ = b c ( a + b ) ( a + c ) Δ \therefore \Delta AC'B' = \frac{bc}{(a + b)(a + c)}\Delta ∴ Δ A C ′ B ′ = ( a + b ) ( a + c ) b c Δ
Similalry, Δ B A ′ C ′ = a c ( a + b ) ( b + c ) Δ \Delta BA'C' = \frac{ac}{(a + b)(b + c)}\Delta Δ B A ′ C ′ = ( a + b ) ( b + c ) a c Δ
and Δ C A ′ B ′ = a b ( a + c ) ( b + c ) Δ \Delta CA'B' = \frac{ab}{(a + c)(b + c)}\Delta Δ C A ′ B ′ = ( a + c ) ( b + c ) ab Δ
∴ Δ A ′ B ′ C ′ = Δ . 2 a b c ( a + b ) ( b + c ) ( c + a ) \therefore \Delta A'B'C' = \Delta.\frac{2abc}{(a + b)(b + c)(c + a)} ∴ Δ A ′ B ′ C ′ = Δ. ( a + b ) ( b + c ) ( c + a ) 2 ab c
∴ Δ A ′ B ′ C ′ Δ A B C = 2 sin A sin B sin C ( sin A + sin B ) ( sin B + sin C ) ( sin C + sin A ) \therefore \frac{\Delta A'B'C'}{\Delta ABC} = \frac{2\sin A\sin B\sin C}{(\sin A + \sin B)(\sin B + \sin C)(\sin C + \sin A)} ∴ Δ A BC Δ A ′ B ′ C ′ = ( s i n A + s i n B ) ( s i n B + s i n C ) ( s i n C + s i n A ) 2 s i n A s i n B s i n C
= 2 sin A 2 sin B 2 sin C 2 cos A − B 2 cos B − C 2 cos C − A 2 = \frac{2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\cos\frac{A - B}{2}\cos\frac{B - C}{2}\cos\frac{C - A}{2}} = c o s 2 A − B c o s 2 B − C c o s 2 C − A 2 s i n 2 A s i n 2 B s i n 2 C
The diagram is given below:
Let the given triangle be A B C ABC A BC and the similar triangle inscribed in triangle A B C ABC A BC be A ′ B ′ C ′ A'B'C' A ′ B ′ C ′ such that
A = A ′ , B = B ′ , C = C ′ A=A', B=B', C=C' A = A ′ , B = B ′ , C = C ′
Let B ′ C ′ = λ a , A ′ C ′ = λ b , A ′ B ′ = λ c B'C'=\lambda a, A'C'=\lambda b, A'B' = \lambda c B ′ C ′ = λa , A ′ C ′ = λb , A ′ B ′ = λ c
According to the quuestion ∠ B ′ O C = θ \angle B'OC=\theta ∠ B ′ OC = θ
Clearly, ∠ O C ′ B = B − θ = ∠ A C ′ B ′ \angle OC'B = B - \theta = \angle AC'B' ∠ O C ′ B = B − θ = ∠ A C ′ B ′
∠ B C ′ A ′ = 18 0 ∘ − ( B − θ + C ) = A + θ \angle BC'A' = 180^\circ - (B - \theta + C) = A + \theta ∠ B C ′ A ′ = 18 0 ∘ − ( B − θ + C ) = A + θ
∠ A B ′ C ′ = 18 0 ∘ − ( B − θ + A ) = C + θ \angle AB'C' = 180^\circ - (B - \theta + A) = C + \theta ∠ A B ′ C ′ = 18 0 ∘ − ( B − θ + A ) = C + θ
∠ A ′ B ′ C ′ = 18 0 ∘ − ( C + θ + B ) = A − θ \angle A'B'C' = 180^\circ - (C + \theta + B) = A - \theta ∠ A ′ B ′ C ′ = 18 0 ∘ − ( C + θ + B ) = A − θ
Applying sine rule in the triangle B C ′ A ′ , BC'A', B C ′ A ′ ,
B A ′ sin ( A + θ ) = λ b sin B ⇒ B A ′ = λ b sin B sin ( A + θ ) \frac{BA'}{\sin(A +\theta)} = \frac{\lambda b}{\sin B}\Rightarrow BA' = \frac{\lambda b}{\sin B}\sin(A + \theta) s i n ( A + θ ) B A ′ = s i n B λb ⇒ B A ′ = s i n B λb sin ( A + θ )
⇒ B A ′ = λ 2 R sin ( A + θ ) \Rightarrow BA' = \lambda 2R\sin(A + \theta) ⇒ B A ′ = λ 2 R sin ( A + θ )
Applying sine rule in A ′ B ′ C , A'B'C, A ′ B ′ C ,
A ′ C sin ( A − θ ) = λ c sin C \frac{A'C}{\sin(A - \theta)} = \frac{\lambda c}{\sin C} s i n ( A − θ ) A ′ C = s i n C λ c
⇒ A ′ C = λ 2 R sin ( A − θ ) \Rightarrow A'C = \lambda 2R\sin(A - \theta) ⇒ A ′ C = λ 2 R sin ( A − θ )
B C = B A ′ + A ′ C BC = BA' + A'C BC = B A ′ + A ′ C
⇒ a = λ 2 R sin ( A + θ ) + λ 2 R sin ( A − θ ) \Rightarrow a = \lambda 2R\sin(A + \theta) + \lambda 2R\sin(A - \theta) ⇒ a = λ 2 R sin ( A + θ ) + λ 2 R sin ( A − θ )
⇒ a = 2 R λ [ sin ( A + θ ) + sin ( A − θ ) ] = 2 R λ 2 sin A cos θ \Rightarrow a = 2R\lambda[\sin(A + \theta) + \sin(A - \theta)] = 2R\lambda 2\sin A\cos\theta ⇒ a = 2 R λ [ sin ( A + θ ) + sin ( A − θ )] = 2 R λ 2 sin A cos θ
⇒ a = 2 λ a cos θ \Rightarrow a = 2\lambda a\cos\theta ⇒ a = 2 λa cos θ
⇒ 2 λ cos θ = 1 \Rightarrow 2\lambda\cos\theta = 1 ⇒ 2 λ cos θ = 1
We have to prove that r 1 + r 2 + r 3 − r = 4 R r_1 + r_2 + r_3 - r = 4R r 1 + r 2 + r 3 − r = 4 R
L.H.S. = Δ s − a + Δ s − b + Δ s − c − Δ s = \frac{\Delta}{s - a} + \frac{\Delta}{s - b} + \frac{\Delta}{s - c} - \frac{\Delta}{s} = s − a Δ + s − b Δ + s − c Δ − s Δ
= Δ [ s − a + s − b ( s − a ) ( s − b ) + s − s + c s ( s − c ) ] = \Delta\left[\frac{s - a + s - b}{(s - a)(s - b)} + \frac{s - s + c}{s(s - c)}\right] = Δ [ ( s − a ) ( s − b ) s − a + s − b + s ( s − c ) s − s + c ]
= Δ [ c ( s − a ) ( s − b ) + c s ( s − c ) ] =\Delta\left[\frac{c}{(s - a)(s - b)} + \frac{c}{s(s - c)}\right] = Δ [ ( s − a ) ( s − b ) c + s ( s − c ) c ]
= Δ . c [ s ( s − c ) + ( s − a ) ( s − b ) s ( s − a ) ( s − b ) ( s − c ) ] =\Delta.c \left[\frac{s(s - c) + (s - a)(s - b)}{s(s - a)(s - b)(s - c)}\right] = Δ. c [ s ( s − a ) ( s − b ) ( s − c ) s ( s − c ) + ( s − a ) ( s − b ) ]
= Δ . c . 1 Δ 2 [ s 2 − s c + s 2 − ( a + b ) s + a b ] =\Delta.c.\frac{1}{\Delta^2}[s^2 - sc + s^2 - (a + b)s + ab] = Δ. c . Δ 2 1 [ s 2 − sc + s 2 − ( a + b ) s + ab ]
= c Δ [ 2 s 2 − s ( a + b + c ) + a b ] = c Δ [ 2 s 2 − 2 s 2 + a b ] =\frac{c}{\Delta}[2s^2 - s(a + b + c) + ab] = \frac{c}{\Delta}[2s^2 - 2s^2 + ab] = Δ c [ 2 s 2 − s ( a + b + c ) + ab ] = Δ c [ 2 s 2 − 2 s 2 + ab ]
= a b c Δ = 4 R = = \frac{abc}{\Delta} = 4R = = Δ ab c = 4 R = R.H.S.
We have to prove that 1 r 1 + 1 r 2 + 1 r 3 = 1 r \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r} r 1 1 + r 2 1 + r 3 1 = r 1
L.H.S. = s − a Δ + s − b Δ + s − c Δ = \frac{s - a}{\Delta} + \frac{s - b}{\Delta} + \frac{s - c}{\Delta} = Δ s − a + Δ s − b + Δ s − c
= 3 s − ( a + b + c ) Δ = 3 s − 2 s Δ [ ∵ 2 s = a + b + c ] = \frac{3s - (a + b + c)}{\Delta} = \frac{3s - 2s}{\Delta} [\because 2s = a + b + c] = Δ 3 s − ( a + b + c ) = Δ 3 s − 2 s [ ∵ 2 s = a + b + c ]
= s Δ = 1 r = = \frac{s}{\Delta} = \frac{1}{r} = = Δ s = r 1 = R.H.S.
We have to prove that 1 r 1 2 + 1 r 2 2 + 1 r 3 2 + 1 r 2 = a 2 + b 2 + c 2 Δ 2 \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2} = \frac{a^2 + b^2 +
c^2}{\Delta^2} r 1 2 1 + r 2 2 1 + r 3 2 1 + r 2 1 = Δ 2 a 2 + b 2 + c 2
L.H.S. = 1 r 1 2 + 1 r 2 2 + 1 r 3 2 + 1 r 2 = \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2} = r 1 2 1 + r 2 2 1 + r 3 2 1 + r 2 1
= ( s − a ) 2 + ( s − b ) 2 + ( s − c ) 2 + s 2 Δ 2 = \frac{(s - a)^2 + (s - b)^2 + (s - c)^2 + s^2}{\Delta^2} = Δ 2 ( s − a ) 2 + ( s − b ) 2 + ( s − c ) 2 + s 2
= 4 s 2 − 2 s ( a + b + c ) + a 2 + b 2 + c 2 Δ 2 = \frac{4s^2 - 2s(a + b + c) + a^2 + b^2 + c^2}{\Delta^2} = Δ 2 4 s 2 − 2 s ( a + b + c ) + a 2 + b 2 + c 2
= 4 s 2 − 2 s . 2 s + a 2 + b 2 + c 2 Δ 2 =\frac{4s^2 - 2s.2s + a^2 + b^2 + c^2}{\Delta^2} = Δ 2 4 s 2 − 2 s .2 s + a 2 + b 2 + c 2
= a 2 + b 2 + c 2 Δ 2 = = \frac{a^2 + b^2 + c^2}{\Delta^2} = = Δ 2 a 2 + b 2 + c 2 = R.H.S.
r = Δ s = Δ s s − a s − a r = \frac{\Delta}{s} = \frac{\Delta}{s}\frac{s - a}{s - a} r = s Δ = s Δ s − a s − a
We know that tan A 2 = ( s − b ) ( s − c ) s ( s − a ) \tan \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} tan 2 A = s ( s − a ) ( s − b ) ( s − c ) and Δ = s ( s − a ) ( s − b ) ( s − c ) \Delta = \sqrt{s(s - a)(s - b)(s -
c)} Δ = s ( s − a ) ( s − b ) ( s − c )
⇒ r = ( s − a ) tan A 3 \Rightarrow r = (s - a)\tan\frac{A}{3} ⇒ r = ( s − a ) tan 3 A
Similarly, r = ( s − b ) tan B 2 , r = ( s − c ) tan C 2 r = (s - b)\tan\frac{B}{2}, r = (s - c)\tan\frac{C}{2} r = ( s − b ) tan 2 B , r = ( s − c ) tan 2 C
We have to prove that 1 A = 1 A 1 + 1 A 2 + 1 A 3 \frac{1}{\sqrt{A}} = \frac{1}{\sqrt{A_1}} + \frac{1}{\sqrt{A_2}} + \frac{1}{\sqrt{A_3}} A 1 = A 1 1 + A 2 1 + A 3 1
R.H.S. = 1 A 1 + 1 A 2 + 1 A 3 = \frac{1}{\sqrt{A_1}} + \frac{1}{\sqrt{A_2}} + \frac{1}{\sqrt{A_3}} = A 1 1 + A 2 1 + A 3 1
= 1 π ( 1 r 1 + 1 r 2 + 1 r 3 ) = \frac{1}{\sqrt{\pi}}\left(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\right) = π 1 ( r 1 1 + r 2 1 + r 3 1 )
= 1 π ( s − a Δ + s − b Δ + s − c Δ ) =\frac{1}{\sqrt{\pi}}\left(\frac{s - a}{\Delta} + \frac{s - b}{\Delta} + \frac{s - c}{\Delta}\right) = π 1 ( Δ s − a + Δ s − b + Δ s − c )
= 1 π ( 3 s − ( a + b + c ) Δ ) = 1 π s Δ = 1 π . 1 r = \frac{1}{\sqrt{\pi}}\left(\frac{3s - (a + b + c)}{\Delta}\right) = \frac{1}{\sqrt{\pi}}\frac{s}{\Delta} =
\frac{1}{\sqrt{\pi}}.\frac{1}{r} = π 1 ( Δ 3 s − ( a + b + c ) ) = π 1 Δ s = π 1 . r 1
= 1 A = = \frac{1}{\sqrt{A}} = = A 1 = L.H.S.
r 1 b c + r 2 c a + r 3 a b = s tan A 2 b c + s tan B 2 c a = s tan C 2 a b \frac{r_1}{bc} + \frac{r_2}{ca} + \frac{r_3}{ab} = \frac{s\tan\frac{A}{2}}{bc} + \frac{s\tan\frac{B}{2}}{ca} =
\frac{s\tan\frac{C}{2}}{ab} b c r 1 + c a r 2 + ab r 3 = b c s t a n 2 A + c a s t a n 2 B = ab s t a n 2 C
= s a b c ( a tan A 2 + b tan B 2 + c tan C 2 ) = \frac{s}{abc}\left(a\tan\frac{A}{2} + b\tan\frac{B}{2} + c\tan\frac{C}{2}\right) = ab c s ( a tan 2 A + b tan 2 B + c tan 2 C )
= s a b c ( 2 R sin A tan A 2 + 2 R sin B tan B 2 + 2 R sin C tan C 2 ) = \frac{s}{abc}\left(2R\sin A\tan\frac{A}{2} + 2R\sin B\tan\frac{B}{2} + 2R\sin C\tan\frac{C}{2}\right) = ab c s ( 2 R sin A tan 2 A + 2 R sin B tan 2 B + 2 R sin C tan 2 C )
= s a b c . 4 R ( sin 2 A 2 + sin 2 B 2 + sin 2 C 2 ) = \frac{s}{abc}.4R\left(\sin^2\frac{A}{2} + \sin^2\frac{B}{2} + \sin^2\frac{C}{2}\right) = ab c s .4 R ( sin 2 2 A + sin 2 2 B + sin 2 2 C )
= s Δ ( 1 − cos A 2 + 1 − cos B 2 + 1 − cos C 2 ) = \frac{s}{\Delta}\left(\frac{1 - \cos A}{2} + \frac{1 - \cos B}{2} + \frac{1 - \cos C}{2}\right) = Δ s ( 2 1 − c o s A + 2 1 − c o s B + 2 1 − c o s C )
= 1 r ( 3 2 − cos A + cos B + cos C 2 ) = \frac{1}{r}\left(\frac{3}{2} - \frac{\cos A + \cos B + \cos C}{2}\right) = r 1 ( 2 3 − 2 c o s A + c o s B + c o s C )
We know that cos A + cos B + cos C = 1 + r R \cos A + \cos B + \cos C = 1 + \frac{r}{R} cos A + cos B + cos C = 1 + R r
= 1 r ( 1 − r 2 R ) = 1 r − 1 2 R = = \frac{1}{r}\left(1 - \frac{r}{2R}\right) = \frac{1}{r} - \frac{1}{2R} = = r 1 ( 1 − 2 R r ) = r 1 − 2 R 1 = R.H.S.
Let D D D be the point where perpendicular from A A A meets B C . BC. BC . Then A D = h AD = h A D = h
The diagram is given below:
Clearly, O B = r , A D = h , O D = h − r OB = r, AD = h, OD=h - r OB = r , A D = h , O D = h − r (If O O O is below
B D BD B D then O D = r − h OD = r - h O D = r − h )
B D = O B 2 − O B 2 = r 2 − ( h − r ) 2 = 2 r h − h 2 BD = \sqrt{OB^2 - OB^2} = \sqrt{r^2 - (h - r)^2} = \sqrt{2rh - h^2} B D = O B 2 − O B 2 = r 2 − ( h − r ) 2 = 2 r h − h 2
Area of triangle = 1 2 . 2. B D . h = h 2 r h − h 2 = \frac{1}{2}.2.BD.h = h\sqrt{2rh - h^2} = 2 1 .2. B D . h = h 2 r h − h 2
Let the sides be a , b , c a, b, c a , b , c then Δ = 1 2 a p 1 = 1 2 b p 2 = 1 2 c p 3 \Delta = \frac{1}{2}ap_1 = \frac{1}{2}bp_2 = \frac{1}{2}cp_3 Δ = 2 1 a p 1 = 2 1 b p 2 = 2 1 c p 3
⇒ p 1 = 2 Δ a , p 2 = 2 Δ b , p 3 = 2 Δ c \Rightarrow p_1 = \frac{2\Delta}{a}, p_2 = \frac{2\Delta}{b}, p_3 = \frac{2\Delta}{c} ⇒ p 1 = a 2Δ , p 2 = b 2Δ , p 3 = c 2Δ
L.H.S. = cos A p 1 + cos B p 2 + cos C p 3 = \frac{\cos A}{p_1} + \frac{\cos B}{p_2} + \frac{\cos C}{p_3} = p 1 c o s A + p 2 c o s B + p 3 c o s C
= 1 2 Δ [ a cos A + b cos B + c cos C ] = \frac{1}{2\Delta}[a\cos A + b\cos B + c\cos C] = 2Δ 1 [ a cos A + b cos B + c cos C ]
= 2 R 2 Δ [ sin A cos A + sin B cos B + sin C cos C ] = \frac{2R}{2\Delta}[\sin A\cos A + \sin B\cos B + \sin C\cos C] = 2Δ 2 R [ sin A cos A + sin B cos B + sin C cos C ]
= R 2 Δ [ sin 2 A + sin 2 B + sin 2 C ] = R 2 Δ 4 sin A sin B sin C = \frac{R}{2\Delta}[\sin 2A + \sin 2B + \sin 2C] = \frac{R}{2\Delta}4\sin A\sin B\sin C = 2Δ R [ sin 2 A + sin 2 B + sin 2 C ] = 2Δ R 4 sin A sin B sin C
= a b c 4 Δ . 1 R 2 = R R 2 = 1 R = = \frac{abc}{4\Delta}.\frac{1}{R^2} = \frac{R}{R^2} = \frac{1}{R} = = 4Δ ab c . R 2 1 = R 2 R = R 1 = R.H.S.
This has been already proved in 149.
L.H.S. = r 1 r 2 r 3 = Δ s − a Δ s − b . Δ s − c = r_1r_2r_3 = \frac{\Delta}{s - a}\frac{\Delta}{s - b}.\frac{\Delta}{s - c} = r 1 r 2 r 3 = s − a Δ s − b Δ . s − c Δ
= Δ 3 ( s − a ) ( s − b ) ) ( s − c ) = Δ . s = \frac{\Delta^3}{(s - a)(s - b))(s - c)} = \Delta .s = ( s − a ) ( s − b )) ( s − c ) Δ 3 = Δ. s
R.H.S. = r 3 cot 2 A 2 cot 2 B 2 cot 2 C 2 = r^3\cot^2\frac{A}{2}\cot^2\frac{B}{2}\cot^2\frac{C}{2} = r 3 cot 2 2 A cot 2 2 B cot 2 2 C
= Δ 3 s 3 . s 2 ( s − a ) 2 Δ 2 . s 2 ( s − b ) 2 Δ 2 . s 2 ( s − c ) 2 Δ 2 = \frac{\Delta^3}{s^3}.\frac{s^2(s - a)^2}{\Delta^2}.\frac{s^2(s - b)^2}{\Delta^2}.\frac{s^2(s - c)^2}{\Delta^2} = s 3 Δ 3 . Δ 2 s 2 ( s − a ) 2 . Δ 2 s 2 ( s − b ) 2 . Δ 2 s 2 ( s − c ) 2
= s 3 ( s − a ) 2 ( s − b ) 2 ( s − c ) 2 Δ 3 = Δ . s = \frac{s^3(s - a)^2(s - b)^2(s - c)^2}{\Delta^3} = \Delta .s = Δ 3 s 3 ( s − a ) 2 ( s − b ) 2 ( s − c ) 2 = Δ. s
We have to prove that a ( r r 1 + r 2 r 3 ) = b ( r r 2 + r 3 r 1 ) = c ( r r 3 + r 1 r 2 ) = a b c a(rr_1 + r2r_3) = b(rr_2 + r_3r_1) = c(rr_3 + r_1r_2) = abc a ( r r 1 + r 2 r 3 ) = b ( r r 2 + r 3 r 1 ) = c ( r r 3 + r 1 r 2 ) = ab c
a ( r r 1 + r 2 r 3 ) = a ( Δ s . Δ s − a + Δ ( s − b ) Δ s − c ) a(rr_1 + r_2r_3) = a\left(\frac{\Delta}{s}.\frac{\Delta}{s - a} + \frac{\Delta}{(s - b)}\frac{\Delta}{s - c}\right) a ( r r 1 + r 2 r 3 ) = a ( s Δ . s − a Δ + ( s − b ) Δ s − c Δ )
= Δ 2 . a ( ( s − b ) ( s − c ) + s ( s − a ) s ( s − a ) ( s − b ) ( s − c ) ) = \Delta^2.a\left(\frac{(s - b)(s - c) + s(s - a)}{s(s - a)(s - b)(s - c)}\right) = Δ 2 . a ( s ( s − a ) ( s − b ) ( s − c ) ( s − b ) ( s − c ) + s ( s − a ) )
= Δ 2 . a ( 2 s 2 − s ( a + b + c ) + b c Δ 2 ) = \Delta^2.a\left(\frac{2s^2 - s(a + b + c) + bc}{\Delta^2}\right) = Δ 2 . a ( Δ 2 2 s 2 − s ( a + b + c ) + b c )
= a b c = abc = ab c
Similalry other terms can be evaluated to same value of a b c . abc. ab c .
We have to prove that ( r 1 + r 2 ) tan C 2 = ( r 3 − r ) cot C 2 = c (r_1 + r_2)\tan\frac{C}{2} = (r_3 - r)\cot\frac{C}{2} = c ( r 1 + r 2 ) tan 2 C = ( r 3 − r ) cot 2 C = c
( r 1 + r 2 ) tan C 2 = ( Δ s − a + Δ s − b ) ( s − a ) ( s − b ) Δ (r_1 + r_2)\tan\frac{C}{2} = \left(\frac{\Delta}{s - a} + \frac{\Delta}{s - b}\right)\frac{(s - a)(s - b)}{\Delta} ( r 1 + r 2 ) tan 2 C = ( s − a Δ + s − b Δ ) Δ ( s − a ) ( s − b )
= s − b + s − a = c = s - b + s - a = c = s − b + s − a = c
( r 3 − r ) cot C 2 = ( Δ s − c − Δ s ) Δ ( s − a ) ( s − b ) (r_3 - r)\cot\frac{C}{2} = \left(\frac{\Delta}{s - c} - \frac{\Delta}{s}\right)\frac{\Delta}{(s - a)( s - b)} ( r 3 − r ) cot 2 C = ( s − c Δ − s Δ ) ( s − a ) ( s − b ) Δ
= Δ 2 ( s − s + c s ( s − a ) ( s − b ) ( s − c ) ) = c = \Delta^2\left(\frac{s - s + c}{s(s - a)(s - b)(s - c)}\right) = c = Δ 2 ( s ( s − a ) ( s − b ) ( s − c ) s − s + c ) = c
We have to prove that 4 R sin A sin B sin C = a cos A + b cos B + c cos C 4R\sin A\sin B\sin C = a\cos A + b\cos B + c\cos C 4 R sin A sin B sin C = a cos A + b cos B + c cos C
R.H.S. = R ( 2 sin A cos A + 2 sin B cos B + 2 sin C cos C ) = R ( sin 2 A + sin 2 B + sin 2 C ) = R(2\sin A\cos A + 2\sin B\cos B + 2\sin C\cos C) = R(\sin 2A + \sin 2B + \sin 2C) = R ( 2 sin A cos A + 2 sin B cos B + 2 sin C cos C ) = R ( sin 2 A + sin 2 B + sin 2 C )
= R ( 2 sin ( A + B ) cos ( A − B ) + 2 sin C cos C ) = 2 R ( sin C cos ( A − B ) + sin C cos C ) [ ∵ sin ( A + B ) = sin ( π − C ) = sin C ] = R(2\sin(A + B)\cos(A - B) + 2\sin C\cos C) = 2R(\sin C\cos(A - B) + \sin C\cos C)[\because \sin(A + B) =
\sin(\pi - C) = \sin C] = R ( 2 sin ( A + B ) cos ( A − B ) + 2 sin C cos C ) = 2 R ( sin C cos ( A − B ) + sin C cos C ) [ ∵ sin ( A + B ) = sin ( π − C ) = sin C ]
= 2 R sin C [ cos ( A − B ) − cos ( A + B ) ] [ ∵ cos C = cos ( π − A − B ) = − cos ( A + B ) ] = 2R\sin C[\cos(A - B) - \cos(A + B)][\because \cos C = \cos(\pi - A - B) = -\cos(A + B)] = 2 R sin C [ cos ( A − B ) − cos ( A + B )] [ ∵ cos C = cos ( π − A − B ) = − cos ( A + B )]
= 2 R sin C . 2 sin A sin B = 4 R sin A sin B sin C = = 2R\sin C. 2\sin A\sin B = 4R\sin A\sin B\sin C = = 2 R sin C .2 sin A sin B = 4 R sin A sin B sin C = L.H.S.
We have to prove that ( r 1 − r ) ( r 2 − r ) ( r 3 − r ) = 4 R r 2 (r_1 - r)(r_2 - r)(r_3 - r) = 4Rr^2 ( r 1 − r ) ( r 2 − r ) ( r 3 − r ) = 4 R r 2
L.H.S. = ( Δ s − a − Δ s ) ( Δ s − b − Δ S ) ( Δ s − c − Δ s ) = \left(\frac{\Delta}{s - a} - \frac{\Delta}{s}\right)\left(\frac{\Delta}{s - b} -
\frac{\Delta}{S}\right)\left(\frac{\Delta}{s - c} - \frac{\Delta}{s}\right) = ( s − a Δ − s Δ ) ( s − b Δ − S Δ ) ( s − c Δ − s Δ )
= Δ 3 ( s − s + a s ( s − a ) ) ( s − s + b s ( s − b ) ) ( s − s + c s ( s − c ) ) = \Delta^3\left(\frac{s - s + a}{s(s - a)}\right)\left(\frac{s - s + b}{s(s - b)}\right)\left(\frac{s - s + c}{s(s -
c)}\right) = Δ 3 ( s ( s − a ) s − s + a ) ( s ( s − b ) s − s + b ) ( s ( s − c ) s − s + c )
= Δ 3 . a b c s 3 ( s − a ) ( s − b ) ( s − c ) = Δ 3 . a b c s 2 Δ 2 = a b c . Δ s 2 = \Delta^3.\frac{abc}{s^3(s - a)(s - b)(s - c)} = \frac{\Delta^3.abc}{s^2\Delta^2} = \frac{abc.\Delta}{s^2} = Δ 3 . s 3 ( s − a ) ( s − b ) ( s − c ) ab c = s 2 Δ 2 Δ 3 . ab c = s 2 ab c .Δ
= a b c Δ . Δ 2 s 2 = 4 R r 2 = = \frac{abc}{\Delta}.\frac{\Delta^2}{s^2} = 4Rr^2 = = Δ ab c . s 2 Δ 2 = 4 R r 2 = R.H.S.
We have to prove that r 2 + r 1 2 + r 2 2 + r 3 2 = 16 R 2 − a 2 − b 2 − c 2 r^2 + r_1^2 + r_2^2 + r_3^2 = 16R^2 - a^2 - b^2 - c^2 r 2 + r 1 2 + r 2 2 + r 3 2 = 16 R 2 − a 2 − b 2 − c 2
( r 1 + r 2 + r 3 − r ) 2 = r 1 2 + r 2 2 + r 3 2 + r 2 − 2 r ( r 1 + r 2 + r 3 ) + 2 ( r 1 r 2 + r 2 r 3 + r 3 r 1 ) (r_1 + r_2 + r_3 - r)^2 = r_1^2 + r_2^2 + r_3^2 + r^2 - 2r(r_1 + r_2 + r_3) + 2(r_1r_2 + r_2r_3 + r_3r_1) ( r 1 + r 2 + r 3 − r ) 2 = r 1 2 + r 2 2 + r 3 2 + r 2 − 2 r ( r 1 + r 2 + r 3 ) + 2 ( r 1 r 2 + r 2 r 3 + r 3 r 1 )
We know that r 1 + r 2 + r 3 − r = 4 R r_1 + r_2 + r_3 - r = 4R r 1 + r 2 + r 3 − r = 4 R and ( r 1 r 2 + r 2 r 3 + r 3 r 1 ) = s 2 (r_1r_2 + r_2r_3 + r_3r_1) = s^2 ( r 1 r 2 + r 2 r 3 + r 3 r 1 ) = s 2
r ( r 1 + r 2 + r 3 ) = Δ s ( Δ s − a + Δ s − b + Δ s − c ) r(r_1 + r_2 + r_3) = \frac{\Delta}{s}\left(\frac{\Delta}{s -a} + \frac{\Delta}{s - b} + \frac{\Delta}{s - c}\right) r ( r 1 + r 2 + r 3 ) = s Δ ( s − a Δ + s − b Δ + s − c Δ )
= Δ 2 s ( s − a ) + Δ 2 s ( s − b ) + Δ 2 s ( s − c ) = − s 2 + ( a b + b c + c a ) = \frac{\Delta^2}{s(s - a)} + \frac{\Delta^2}{s(s - b)} + \frac{\Delta^2}{s(s - c)} = -s^2 + (ab + bc + ca) = s ( s − a ) Δ 2 + s ( s − b ) Δ 2 + s ( s − c ) Δ 2 = − s 2 + ( ab + b c + c a )
16 R 2 = r 1 2 + r 2 2 + r 3 2 + r 2 − 2 [ − s 2 + ( a b + b c + c a ) ] + 2 s 2 16R^2 = r_1^2 + r_2^2 + r_3^2 + r^2 -2[-s^2 + (ab + bc + ca)] + 2s^2 16 R 2 = r 1 2 + r 2 2 + r 3 2 + r 2 − 2 [ − s 2 + ( ab + b c + c a )] + 2 s 2
⇒ r 2 + r 1 2 + r 2 2 + r 3 2 = 16 R 2 − a 2 − b 2 − c 2 \Rightarrow r^2 + r_1^2 + r_2^2 + r_3^2 = 16R^2 - a^2 - b^2 - c^2 ⇒ r 2 + r 1 2 + r 2 2 + r 3 2 = 16 R 2 − a 2 − b 2 − c 2
We know that I A = r cosec A 2 , I B = r cosec B 2 , I C = r cosec C 2 IA = r\cosec\frac{A}{2}, IB=r\cosec\frac{B}{2}, IC=r\cosec\frac{C}{2} I A = r cosec 2 A , I B = r cosec 2 B , I C = r cosec 2 C
L.H.S. = r 3 sin A 2 sin B 2 sin C 2 = r 3 . 4 R 4 R sin A 2 sin B 2 sin C 2 = \frac{r^3}{\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}} =
\frac{r^3.4R}{4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}} = s i n 2 A s i n 2 B s i n 2 C r 3 = 4 R s i n 2 A s i n 2 B s i n 2 C r 3 .4 R
= r 3 . 4 R r = 4 R . r 2 = 4 R . Δ 2 s 2 = a b c Δ s 2 = \frac{r^3.4R}{r} = 4R.r^2 = 4R.\frac{\Delta^2}{s^2} = \frac{abc\Delta}{s^2} = r r 3 .4 R = 4 R . r 2 = 4 R . s 2 Δ 2 = s 2 ab c Δ
R.H.S. = a b c . ( s − a ) 2 ( s − b ) 2 ( s − c ) 2 Δ 3 = a b c Δ s 2 = abc.\frac{(s - a)^2(s - b)^2(s - c)^2}{\Delta^3} = \frac{abc\Delta}{s^2} = ab c . Δ 3 ( s − a ) 2 ( s − b ) 2 ( s − c ) 2 = s 2 ab c Δ
From here we can
say that A I 1 = r 1 cosec A 2 AI_1 = r_1\cosec\frac{A}{2} A I 1 = r 1 cosec 2 A
I I 1 = A I 1 − A I = r 1 cosec A 2 − r cosec A 2 II_1 = AI_1 - AI = r_1\cosec\frac{A}{2} - r\cosec\frac{A}{2} I I 1 = A I 1 − A I = r 1 cosec 2 A − r cosec 2 A
= ( Δ s − a − Δ s ) cosec A 2 = \left(\frac{\Delta}{s - a} - \frac{\Delta}{s}\right)\cosec\frac{A}{2} = ( s − a Δ − s Δ ) cosec 2 A
= Δ . a s ( s − a ) b c ( s − b ) ( s − c ) = a sec A 3 = \Delta .\frac{a}{s(s - a)}\sqrt{\frac{bc}{(s - b)(s - c)}} = a\sec\frac{A}{3} = Δ. s ( s − a ) a ( s − b ) ( s − c ) b c = a sec 3 A
If E 2 E_2 E 2 be the point of contact of the circle whose center is I 2 I_2 I 2 with the side A C AC A C of the triangle
A B C , ABC, A BC , we have
A I 2 = A E 2 sec I 2 A E 2 = A E 2 s e c ( 9 0 ∘ − A 2 ) = ( s − b ) cosec A 2 AI_2 = AE_2\sec I_2AE_2 = AE_2sec\left(90^\circ - \frac{A}{2}\right) = (s - b)\cosec \frac{A}{2} A I 2 = A E 2 sec I 2 A E 2 = A E 2 sec ( 9 0 ∘ − 2 A ) = ( s − b ) cosec 2 A
I 2 I 3 = A I 2 + A I 3 = ( s − b + s − c ) cosec A 2 = a cosec A 2 I_2I_3 = AI_2 + AI_3 = (s - b + s - c)\cosec\frac{A}{2} = a\cosec\frac{A}{2} I 2