# 21. Properties of Triangles’ Solutions Part 4¶

1. The diagram is given below:

Let $ABC$ be the triangle. Let $O$ be the circumcenter and $I,$ the incenter.

Clearly, $OA = OB = OC = R, IE=r[IE\perp AB]$

Let $OM\perp BC$ then $\angle BOM = \angle COM = A$

Now, $OA = R, AI = r\cosec\frac{A}{2} = \frac{4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\sin\frac{A}{2}}$

$= 4R\sin\frac{B}{2}\sin\frac{C}{2}$

$\angle OAB = \angle OBA = B - (90^\circ - A) = A + B - 90^\circ = 90^\circ - C$

$\therefore \angle OAI = \angle BAI - \angle BAO = \frac{A}{2} - (90^\circ - C)$

$= \frac{C - B}{2}$

Applying cosine law in $\triangle OAI,$

$\cos\frac{C - B}{2} = \frac{OA^2 + AI^2 - OI^2}{2OA.AI}$

$= \frac{R^2 + 16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} - OI^2}{2.R.4R\sin\frac{B}{2}\sin\frac{C}{2}}$

$OI^2 = R^2\left[1 + 8\sin\frac{B}{2}\sin\frac{C}{2}\left\{2\sin\frac{B}{2}\sin\frac{C}{2} - \cos\left(\frac{B - C}{2}\right)\right\}\right]$

$= R^2\left[1 + 8\sin\frac{B}{2}\sin\frac{C}{2}\left\{\cos\frac{B - C}{2} - \cos\frac{B + C}{2} - \cos\frac{B - C}{2}\right\}\right]$

$= R^2\left[1 - 8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right]$

$= R^2 - 2Rr$

Let us prove the necessary condition for the second part.

Let $b$ be the A.M. of $a$ and $c$ i.e. $2b = a + c$

$\Rightarrow 2\sin B = \sin A + \sin C \Rightarrow 2.2\sin\frac{B}{2}\cos\frac{B}{2} = 2\sin\frac{A + C}{2}\cos\frac{A - C}{2}$

$\Rightarrow 2\sin\frac{B}{2} = \cos\frac{A - C}{2} \Rightarrow 2\cos\frac{A + C}{2} = \cos\frac{A - C}{2}$

$\cos BIO = \frac{BI^2 + IO^2 - BO^2}{2BI.IO}$

Now $BI^2 + IO^2 - BO^2 = r^2\cosec^2\frac{B}{2} + R^2 - 2rR - R^2$

$= r^2\cosec^2\frac{B}{2} - 2rR$

$= \frac{r.4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\sin^2\frac{B}{2}} - 2rR$

$= \frac{2rR.2\sin\frac{A}{2}\sin\frac{C}{2}}{\sin\frac{B}{2}} - 2rR$

$= \frac{2rR\left[\cos\frac{A - C}{2} - \cos\frac{A + C}{2}\right]}{\sin\frac{B}{2}} - 2rR$

$= 2rR\frac{2\cos\frac{A + C}{2}- \cos\frac{A + C}{2}}{\sin\frac{B}{2}} - 2rR = 0$

Thus, $\triangle BIO$ is a right angled triangle.

Now the sufficient condition can be proved similarly.

2. $\frac{A}{2} + \frac{B}{2} = \frac{\pi}{2} - \frac{C}{2}$

$\Rightarrow \cot\left(\frac{A}{2} + \frac{B}{2}\right) = \cot\left(\frac{\pi}{2} - \frac{C}{2}\right)$

$\Rightarrow \frac{\cot\frac{A}{2}\cot\frac{B}{2} - 1}{\cot\frac{A}{2} + \cot\frac{B}{2}} = \tan\frac{C}{2} = \frac{1}{\cot\frac{C}{2}}$

$\Rightarrow \cot\frac{A}{2} + \cot \frac{B}{2} + \cot\frac{C}{2} = \cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}$

3. The diagram is given below:

$HL = r_2, ID=r \therefore IL = ID - LD = ID - HK = r - r_2$

In $\triangle IHL,$

$\cot\frac{B}{2} = \frac{HL}{IL} = \frac{r_2}{r - r_2}$

Similarly, $\cot\frac{A}{2} = \frac{r_1}{r - r_1}$ and

$\cot\frac{C}{2} = \frac{r_3}{r - r_3}$

Now following the result of previous problem

$\frac{r_1}{r - r_1} + \frac{r_2}{r - r_2} + \frac{r_3}{r - r_3} = \frac{r_1r_2r_3}{(r - r_1)(r - r_2)(r - r_3)}$

4. The diagram is given below:

Let $O$ be the circumcenter and $P$ the orthocenter of the $\triangle ABC.$ From geometry,

$\angle BOF = \angle COF = A$ and $AP = 2OF = 2R\cos A, BF = R\sin A$

From right angled $\triangle ADB$

$BD = c\cos B, AD = c\sin B$

Now, $PM = AD - (AP + MD) = c\sin B - (2R\cos A + R\cos A) = c\sin B - 3R\cos A$

$OM = FD = BF - BD = R\sin A - c\cos B$

$\therefore \tan\theta = \frac{PM}{OM} = \frac{c\sin B - 3R\cos A}{R\sin A - c\cos B}$

$= \frac{2R\sin C\sin B - 3R\cos A}{R\sin A - 2R\sin C\cos B}$

$= \frac{2\sin B\sin C - 2\cos[\pi - (B + C)]}{\sin[\pi - (B + C)- 2\sin C\cos B]}$

$= \frac{2\sin B\sin C + 3\cos(B + C)}{\sin(B + C) - 2\sin C\cos B}$

$= \frac{3\cos B\cos C - \sin B\sin C}{\cos C\sin B - \sin C\cos B}$

$= \frac{3 - \tan B\tan C}{\tan B - \tan C}$

5. The diagram is given below:

Let $ABC$ be the triangle. Let $O$ and $I$ be the circumcenter and in-center of the $\triangle ABC.$ Let $P$ be the center and $x,$ the radius of the circle drawn which touches the inscribed and circumscribed circle, of $\triangle ABC$ and the side $BC$ externally. Let us join $OP$ and extend up to $Q.$ Let $ID\perp BC.$ Clearly, $P$ will lie on the extended part of $ID.$ Draw the line $ON$ parallel to the line $IP.$ Join $NP.$

Clearly, $OB = OC OQ = R, OP = OQ - PQ = R - x$

$ON = OM + MN = R\cos A + x$

$NP = MD = OM - CD = \frac{a}{2} - r\cot \frac{C}{2}$

$= \frac{a}{2} - \frac{\Delta}{s}.\frac{s(s - c)}{\Delta} = \frac{a}{2} - (s - c)$

$= \frac{c - b}{2}$

From right angled $\triangle ONP, OP^2 = ON^2 + NP^2$

$(R - x)^2 = (R\cos A + x)^2 + \left(c - b\right)^2$

$R^2 + x^2 - 2Rx = R^2\cos^2A + x^2 + 2Rx\cos A + \left(\frac{c - b}{2}\right)^2$

$2Rx(1 + \cos A) = R^2(1 - \cos^2A) - \left(\frac{c - b}{2}\right)^2$

$4Rx\cos^2\frac{A}{2} = R^2\sin^2A - \left(\frac{b - c}{2}\right)^2 = \frac{a^2}{4} - \frac{(b - c)^2}{4}$

$4Rx\frac{s(s - a)}{bc} = \frac{a +b - c}{2}.\frac{a - b + c}{2} = (s - b)(s - c)$

$x = \frac{\Delta}{a}\tan^2\frac{A}{2}$

6. The diagram is given below:

Since angles n the same segment of a circle are equal. $\therefore \angle BED= \angle BAD = \frac{A}{2}$

and $\angle BEF = \angle BCF = \frac{C}{2}$

Now $\angle DEF = \angle BEF + \angle BED = \frac{C}{2} + \frac{A}{2} = \frac{C + A}{2} = 90^\circ - \frac{B}{2}$

Similarly, $DFE = 90^\circ - \frac{C}{2}, \angle EDF = 90^\circ - \frac{A}{2}$

Now area of $\triangle DEF = \frac{1}{2}DE.DF.\sin \angle EDF$

Let $R$ be the circum-radius of $\triangle ABC$ then clearly $R$ is also the circum-radius of $\triangle DEF.$

Applying sine rule in $\triangle DEF,$ we have

$\frac{DE}{\sin DFE} = \frac{DF}{\sin DEF} = \frac{EF}{\sin EDF} = 2R$

$\Rightarrow DE = 2R\sin DFE = 2R\sin\left(90^\circ - \frac{C}{2}\right) = 2R\cos\frac{C}{2}$

Similarly, $DF = 2R\cos\frac{B}{2}$

So, area of $\triangle DEF = \frac{1}{2}.4R^2\cos\frac{B}{2}\cos\frac{C}{2}.\sin\left(90^\circ - \frac{A}{2}\right)$

$= 2R^2\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$

$= 2R^2\sqrt{\frac{s(s - a)}{bc}}\sqrt{\frac{s(s - b)}{ca}}\sqrt{\frac{s(s - c)}{ab}}$

$= \frac{2R^2s\sqrt{s(s - a)(s - b)(s - c)}}{abc} = \frac{2R^2s\Delta}{abc} = \frac{R^2.s.4\Delta}{2abc} = \frac{R^2s}{2\frac{abc}{4\Delta}}$

Here, are of $\triangle ABC = \Delta$

$= \frac{R^2s}{2R} = \frac{R}{2}s$

$\Rightarrow \frac{\Delta DEF}{\Delta ABC} = \frac{Rs}{2\Delta} = \frac{R}{2r}$

7. The diagram is given below:

$\Delta A'B'C' = \Delta ABC - (\Delta AC'B' + \Delta BA'C' + \Delta CA'B')$

$\Delta AC'B' = \frac{1}{2}AC'.AB'\sin A$

$\because CC'$ is the internal bisector of $\angle C$

$\frac{AC'}{C'B} = \frac{AC}{CB} = \frac{b}{a}$

$\Rightarrow AC' = bk, C'B = ak,$ where $k$ is some constant dependent on angles.

$AC' + C'B = AB \Rightarrow c = ak + bk \Rightarrow k = \frac{c}{a + b}$

$\therefore AC' = \frac{bc}{a + b}$

Similarly, $AB' = \frac{bc}{a + c}$

$\Delta AC'B' = \frac{1}{2}\frac{bc}{a + b}\frac{bc}{a + c}\sin A$

Let $\Delta$ be the area of $\triangle ABC,$ then $\Delta = \frac{1}{2}bc\sin A$

$\therefore \Delta AC'B' = \frac{bc}{(a + b)(a + c)}\Delta$

Similalry, $\Delta BA'C' = \frac{ac}{(a + b)(b + c)}\Delta$

and $\Delta CA'B' = \frac{ab}{(a + c)(b + c)}\Delta$

$\therefore \Delta A'B'C' = \Delta.\frac{2abc}{(a + b)(b + c)(c + a)}$

$\therefore \frac{\Delta A'B'C'}{\Delta ABC} = \frac{2\sin A\sin B\sin C}{(\sin A + \sin B)(\sin B + \sin C)(\sin C + \sin A)}$

$= \frac{2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\cos\frac{A - B}{2}\cos\frac{B - C}{2}\cos\frac{C - A}{2}}$

8. The diagram is given below:

Let the given triangle be $ABC$ and the similar triangle inscribed in triangle $ABC$ be $A'B'C'$ such that

$A=A', B=B', C=C'$

Let $B'C'=\lambda a, A'C'=\lambda b, A'B' = \lambda c$

According to the quuestion $\angle B'OC=\theta$

Clearly, $\angle OC'B = B - \theta = \angle AC'B'$

$\angle BC'A' = 180^\circ - (B - \theta + C) = A + \theta$

$\angle AB'C' = 180^\circ - (B - \theta + A) = C + \theta$

$\angle A'B'C' = 180^\circ - (C + \theta + B) = A - \theta$

Applying sine rule in the triangle $BC'A',$

$\frac{BA'}{\sin(A +\theta)} = \frac{\lambda b}{\sin B}\Rightarrow BA' = \frac{\lambda b}{\sin B}\sin(A + \theta)$

$\Rightarrow BA' = \lambda 2R\sin(A + \theta)$

Applying sine rule in $A'B'C,$

$\frac{A'C}{\sin(A - \theta)} = \frac{\lambda c}{\sin C}$

$\Rightarrow A'C = \lambda 2R\sin(A - \theta)$

$BC = BA' + A'C$

$\Rightarrow a = \lambda 2R\sin(A + \theta) + \lambda 2R\sin(A - \theta)$

$\Rightarrow a = 2R\lambda[\sin(A + \theta) + \sin(A - \theta)] = 2R\lambda 2\sin A\cos\theta$

$\Rightarrow a = 2\lambda a\cos\theta$

$\Rightarrow 2\lambda\cos\theta = 1$

9. We have to prove that $r_1 + r_2 + r_3 - r = 4R$

L.H.S. $= \frac{\Delta}{s - a} + \frac{\Delta}{s - b} + \frac{\Delta}{s - c} - \frac{\Delta}{s}$

$= \Delta\left[\frac{s - a + s - b}{(s - a)(s - b)} + \frac{s - s + c}{s(s - c)}\right]$

$=\Delta\left[\frac{c}{(s - a)(s - b)} + \frac{c}{s(s - c)}\right]$

$=\Delta.c \left[\frac{s(s - c) + (s - a)(s - b)}{s(s - a)(s - b)(s - c)}\right]$

$=\Delta.c.\frac{1}{\Delta^2}[s^2 - sc + s^2 - (a + b)s + ab]$

$=\frac{c}{\Delta}[2s^2 - s(a + b + c) + ab] = \frac{c}{\Delta}[2s^2 - 2s^2 + ab]$

$= \frac{abc}{\Delta} = 4R =$ R.H.S.

10. We have to prove that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$

L.H.S. $= \frac{s - a}{\Delta} + \frac{s - b}{\Delta} + \frac{s - c}{\Delta}$

$= \frac{3s - (a + b + c)}{\Delta} = \frac{3s - 2s}{\Delta} [\because 2s = a + b + c]$

$= \frac{s}{\Delta} = \frac{1}{r} =$ R.H.S.

11. We have to prove that $\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2} = \frac{a^2 + b^2 + c^2}{\Delta^2}$

L.H.S. $= \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2}$

$= \frac{(s - a)^2 + (s - b)^2 + (s - c)^2 + s^2}{\Delta^2}$

$= \frac{4s^2 - 2s(a + b + c) + a^2 + b^2 + c^2}{\Delta^2}$

$=\frac{4s^2 - 2s.2s + a^2 + b^2 + c^2}{\Delta^2}$

$= \frac{a^2 + b^2 + c^2}{\Delta^2} =$ R.H.S.

12. $r = \frac{\Delta}{s} = \frac{\Delta}{s}\frac{s - a}{s - a}$

We know that $\tan \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}}$ and $\Delta = \sqrt{s(s - a)(s - b)(s - c)}$

$\Rightarrow r = (s - a)\tan\frac{A}{3}$

Similarly, $r = (s - b)\tan\frac{B}{2}, r = (s - c)\tan\frac{C}{2}$

13. We have to prove that $\frac{1}{\sqrt{A}} = \frac{1}{\sqrt{A_1}} + \frac{1}{\sqrt{A_2}} + \frac{1}{\sqrt{A_3}}$

R.H.S. $= \frac{1}{\sqrt{A_1}} + \frac{1}{\sqrt{A_2}} + \frac{1}{\sqrt{A_3}}$

$= \frac{1}{\sqrt{\pi}}\left(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\right)$

$=\frac{1}{\sqrt{\pi}}\left(\frac{s - a}{\Delta} + \frac{s - b}{\Delta} + \frac{s - c}{\Delta}\right)$

$= \frac{1}{\sqrt{\pi}}\left(\frac{3s - (a + b + c)}{\Delta}\right) = \frac{1}{\sqrt{\pi}}\frac{s}{\Delta} = \frac{1}{\sqrt{\pi}}.\frac{1}{r}$

$= \frac{1}{\sqrt{A}} =$ L.H.S.

14. $\frac{r_1}{bc} + \frac{r_2}{ca} + \frac{r_3}{ab} = \frac{s\tan\frac{A}{2}}{bc} + \frac{s\tan\frac{B}{2}}{ca} = \frac{s\tan\frac{C}{2}}{ab}$

$= \frac{s}{abc}\left(a\tan\frac{A}{2} + b\tan\frac{B}{2} + c\tan\frac{C}{2}\right)$

$= \frac{s}{abc}\left(2R\sin A\tan\frac{A}{2} + 2R\sin B\tan\frac{B}{2} + 2R\sin C\tan\frac{C}{2}\right)$

$= \frac{s}{abc}.4R\left(\sin^2\frac{A}{2} + \sin^2\frac{B}{2} + \sin^2\frac{C}{2}\right)$

$= \frac{s}{\Delta}\left(\frac{1 - \cos A}{2} + \frac{1 - \cos B}{2} + \frac{1 - \cos C}{2}\right)$

$= \frac{1}{r}\left(\frac{3}{2} - \frac{\cos A + \cos B + \cos C}{2}\right)$

We know that $\cos A + \cos B + \cos C = 1 + \frac{r}{R}$

$= \frac{1}{r}\left(1 - \frac{r}{2R}\right) = \frac{1}{r} - \frac{1}{2R} =$ R.H.S.

15. Let $D$ be the point where perpendicular from $A$ meets $BC.$ Then $AD = h$

The diagram is given below:

Clearly, $OB = r, AD = h, OD=h - r$ (If $O$ is below $BD$ then $OD = r - h$)

$BD = \sqrt{OB^2 - OB^2} = \sqrt{r^2 - (h - r)^2} = \sqrt{2rh - h^2}$

Area of triangle $= \frac{1}{2}.2.BD.h = h\sqrt{2rh - h^2}$

16. Let the sides be $a, b, c$ then $\Delta = \frac{1}{2}ap_1 = \frac{1}{2}bp_2 = \frac{1}{2}cp_3$

$\Rightarrow p_1 = \frac{2\Delta}{a}, p_2 = \frac{2\Delta}{b}, p_3 = \frac{2\Delta}{c}$

L.H.S. $= \frac{\cos A}{p_1} + \frac{\cos B}{p_2} + \frac{\cos C}{p_3}$

$= \frac{1}{2\Delta}[a\cos A + b\cos B + c\cos C]$

$= \frac{2R}{2\Delta}[\sin A\cos A + \sin B\cos B + \sin C\cos C]$

$= \frac{R}{2\Delta}[\sin 2A + \sin 2B + \sin 2C] = \frac{R}{2\Delta}4\sin A\sin B\sin C$

$= \frac{abc}{4\Delta}.\frac{1}{R^2} = \frac{R}{R^2} = \frac{1}{R} =$ R.H.S.

17. This has been already proved in 149.

18. L.H.S. $= r_1r_2r_3 = \frac{\Delta}{s - a}\frac{\Delta}{s - b}.\frac{\Delta}{s - c}$

$= \frac{\Delta^3}{(s - a)(s - b))(s - c)} = \Delta .s$

R.H.S. $= r^3\cot^2\frac{A}{2}\cot^2\frac{B}{2}\cot^2\frac{C}{2}$

$= \frac{\Delta^3}{s^3}.\frac{s^2(s - a)^2}{\Delta^2}.\frac{s^2(s - b)^2}{\Delta^2}.\frac{s^2(s - c)^2}{\Delta^2}$

$= \frac{s^3(s - a)^2(s - b)^2(s - c)^2}{\Delta^3} = \Delta .s$

19. We have to prove that $a(rr_1 + r2r_3) = b(rr_2 + r_3r_1) = c(rr_3 + r_1r_2) = abc$

$a(rr_1 + r_2r_3) = a\left(\frac{\Delta}{s}.\frac{\Delta}{s - a} + \frac{\Delta}{(s - b)}\frac{\Delta}{s - c}\right)$

$= \Delta^2.a\left(\frac{(s - b)(s - c) + s(s - a)}{s(s - a)(s - b)(s - c)}\right)$

$= \Delta^2.a\left(\frac{2s^2 - s(a + b + c) + bc}{\Delta^2}\right)$

$= abc$

Similalry other terms can be evaluated to same value of $abc.$

20. We have to prove that $(r_1 + r_2)\tan\frac{C}{2} = (r_3 - r)\cot\frac{C}{2} = c$

$(r_1 + r_2)\tan\frac{C}{2} = \left(\frac{\Delta}{s - a} + \frac{\Delta}{s - b}\right)\frac{(s - a)(s - b)}{\Delta}$

$= s - b + s - a = c$

$(r_3 - r)\cot\frac{C}{2} = \left(\frac{\Delta}{s - c} - \frac{\Delta}{s}\right)\frac{\Delta}{(s - a)( s - b)}$

$= \Delta^2\left(\frac{s - s + c}{s(s - a)(s - b)(s - c)}\right) = c$

21. We have to prove that $4R\sin A\sin B\sin C = a\cos A + b\cos B + c\cos C$

R.H.S. $= R(2\sin A\cos A + 2\sin B\cos B + 2\sin C\cos C) = R(\sin 2A + \sin 2B + \sin 2C)$

$= R(2\sin(A + B)\cos(A - B) + 2\sin C\cos C) = 2R(\sin C\cos(A - B) + \sin C\cos C)[\because \sin(A + B) = \sin(\pi - C) = \sin C]$

$= 2R\sin C[\cos(A - B) - \cos(A + B)][\because \cos C = \cos(\pi - A - B) = -\cos(A + B)]$

$= 2R\sin C. 2\sin A\sin B = 4R\sin A\sin B\sin C =$ L.H.S.

22. We have to prove that $(r_1 - r)(r_2 - r)(r_3 - r) = 4Rr^2$

L.H.S. $= \left(\frac{\Delta}{s - a} - \frac{\Delta}{s}\right)\left(\frac{\Delta}{s - b} - \frac{\Delta}{S}\right)\left(\frac{\Delta}{s - c} - \frac{\Delta}{s}\right)$

$= \Delta^3\left(\frac{s - s + a}{s(s - a)}\right)\left(\frac{s - s + b}{s(s - b)}\right)\left(\frac{s - s + c}{s(s - c)}\right)$

$= \Delta^3.\frac{abc}{s^3(s - a)(s - b)(s - c)} = \frac{\Delta^3.abc}{s^2\Delta^2} = \frac{abc.\Delta}{s^2}$

$= \frac{abc}{\Delta}.\frac{\Delta^2}{s^2} = 4Rr^2 =$ R.H.S.

23. We have to prove that $r^2 + r_1^2 + r_2^2 + r_3^2 = 16R^2 - a^2 - b^2 - c^2$

$(r_1 + r_2 + r_3 - r)^2 = r_1^2 + r_2^2 + r_3^2 + r^2 - 2r(r_1 + r_2 + r_3) + 2(r_1r_2 + r_2r_3 + r_3r_1)$

We know that $r_1 + r_2 + r_3 - r = 4R$ and $(r_1r_2 + r_2r_3 + r_3r_1) = s^2$

$r(r_1 + r_2 + r_3) = \frac{\Delta}{s}\left(\frac{\Delta}{s -a} + \frac{\Delta}{s - b} + \frac{\Delta}{s - c}\right)$

$= \frac{\Delta^2}{s(s - a)} + \frac{\Delta^2}{s(s - b)} + \frac{\Delta^2}{s(s - c)} = -s^2 + (ab + bc + ca)$

$16R^2 = r_1^2 + r_2^2 + r_3^2 + r^2 -2[-s^2 + (ab + bc + ca)] + 2s^2$

$\Rightarrow r^2 + r_1^2 + r_2^2 + r_3^2 = 16R^2 - a^2 - b^2 - c^2$

24. We know that $IA = r\cosec\frac{A}{2}, IB=r\cosec\frac{B}{2}, IC=r\cosec\frac{C}{2}$

L.H.S. $= \frac{r^3}{\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}} = \frac{r^3.4R}{4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}$

$= \frac{r^3.4R}{r} = 4R.r^2 = 4R.\frac{\Delta^2}{s^2} = \frac{abc\Delta}{s^2}$

R.H.S. $= abc.\frac{(s - a)^2(s - b)^2(s - c)^2}{\Delta^3} = \frac{abc\Delta}{s^2}$

25. From here we can say that $AI_1 = r_1\cosec\frac{A}{2}$

26. $II_1 = AI_1 - AI = r_1\cosec\frac{A}{2} - r\cosec\frac{A}{2}$

$= \left(\frac{\Delta}{s - a} - \frac{\Delta}{s}\right)\cosec\frac{A}{2}$

$= \Delta .\frac{a}{s(s - a)}\sqrt{\frac{bc}{(s - b)(s - c)}} = a\sec\frac{A}{3}$

27. If $E_2$ be the point of contact of the circle whose center is $I_2$ with the side $AC$ of the triangle $ABC,$ we have

$AI_2 = AE_2\sec I_2AE_2 = AE_2sec\left(90^\circ - \frac{A}{2}\right) = (s - b)\cosec \frac{A}{2}$

$I_2I_3 = AI_2 + AI_3 = (s - b + s - c)\cosec\frac{A}{2} = a\cosec\frac{A}{2}$