21. Properties of Triangles’ Solutions Part 4

  1. The diagram is given below:

    Problem 151

    Let ABCABC be the triangle. Let OO be the circumcenter and I,I, the incenter.

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

    Let OMBCOM\perp BC then BOM=COM=A\angle BOM = \angle COM = A

    Now, OA=R,AI=rcosecA2=4RsinA2sinB2sinC2sinA2OA = 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}}

    =4RsinB2sinC2= 4R\sin\frac{B}{2}\sin\frac{C}{2}

    OAB=OBA=B(90A)=A+B90=90C\angle OAB = \angle OBA = B - (90^\circ - A) = A + B - 90^\circ = 90^\circ - C

    OAI=BAIBAO=A2(90C)\therefore \angle OAI = \angle BAI - \angle BAO = \frac{A}{2} - (90^\circ - C)

    =CB2= \frac{C - B}{2}

    Applying cosine law in OAI,\triangle OAI,

    cosCB2=OA2+AI2OI22OA.AI\cos\frac{C - B}{2} = \frac{OA^2 + AI^2 - OI^2}{2OA.AI}

    =R2+16R2sin2B2sin2C2OI22.R.4RsinB2sinC2= \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}}

    OI2=R2[1+8sinB2sinC2{2sinB2sinC2cos(BC2)}]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]

    =R2[1+8sinB2sinC2{cosBC2cosB+C2cosBC2}]= 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]

    =R2[18sinA2sinB2sinC2]= R^2\left[1 - 8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right]

    =R22Rr= R^2 - 2Rr

    Let us prove the necessary condition for the second part.

    Let bb be the A.M. of aa and cc i.e. 2b=a+c2b = a + c

    2sinB=sinA+sinC2.2sinB2cosB2=2sinA+C2cosAC2\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}

    2sinB2=cosAC22cosA+C2=cosAC2\Rightarrow 2\sin\frac{B}{2} = \cos\frac{A - C}{2} \Rightarrow 2\cos\frac{A + C}{2} = \cos\frac{A - C}{2}

    cosBIO=BI2+IO2BO22BI.IO\cos BIO = \frac{BI^2 + IO^2 - BO^2}{2BI.IO}

    Now BI2+IO2BO2=r2cosec2B2+R22rRR2BI^2 + IO^2 - BO^2 = r^2\cosec^2\frac{B}{2} + R^2 - 2rR - R^2

    =r2cosec2B22rR= r^2\cosec^2\frac{B}{2} - 2rR

    =r.4RsinA2sinB2sinC2sin2B22rR= \frac{r.4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\sin^2\frac{B}{2}} - 2rR

    =2rR.2sinA2sinC2sinB22rR= \frac{2rR.2\sin\frac{A}{2}\sin\frac{C}{2}}{\sin\frac{B}{2}} - 2rR

    =2rR[cosAC2cosA+C2]sinB22rR= \frac{2rR\left[\cos\frac{A - C}{2} - \cos\frac{A + C}{2}\right]}{\sin\frac{B}{2}} - 2rR

    =2rR2cosA+C2cosA+C2sinB22rR=0= 2rR\frac{2\cos\frac{A + C}{2}- \cos\frac{A + C}{2}}{\sin\frac{B}{2}} - 2rR = 0

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

    Now the sufficient condition can be proved similarly.

  2. A2+B2=π2C2\frac{A}{2} + \frac{B}{2} = \frac{\pi}{2} - \frac{C}{2}

    cot(A2+B2)=cot(π2C2)\Rightarrow \cot\left(\frac{A}{2} + \frac{B}{2}\right) = \cot\left(\frac{\pi}{2} - \frac{C}{2}\right)

    cotA2cotB21cotA2+cotB2=tanC2=1cotC2\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}}

    cotA2+cotB2+cotC2=cotA2cotB2cotC2\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:

    Problem 153

    HL=r2,ID=rIL=IDLD=IDHK=rr2HL = r_2, ID=r \therefore IL = ID - LD = ID - HK = r - r_2

    In IHL,\triangle IHL,

    cotB2=HLIL=r2rr2\cot\frac{B}{2} = \frac{HL}{IL} = \frac{r_2}{r - r_2}

    Similarly, cotA2=r1rr1\cot\frac{A}{2} = \frac{r_1}{r - r_1} and

    cotC2=r3rr3\cot\frac{C}{2} = \frac{r_3}{r - r_3}

    Now following the result of previous problem

    r1rr1+r2rr2+r3rr3=r1r2r3(rr1)(rr2)(rr3)\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:

    Problem 154

    Let OO be the circumcenter and PP the orthocenter of the ABC.\triangle ABC. From geometry,

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

    From right angled ADB\triangle ADB

    BD=ccosB,AD=csinBBD = c\cos B, AD = c\sin B

    Now, PM=AD(AP+MD)=csinB(2RcosA+RcosA)=csinB3RcosAPM = AD - (AP + MD) = c\sin B - (2R\cos A + R\cos A) = c\sin B - 3R\cos A

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

    tanθ=PMOM=csinB3RcosARsinAccosB\therefore \tan\theta = \frac{PM}{OM} = \frac{c\sin B - 3R\cos A}{R\sin A - c\cos B}

    =2RsinCsinB3RcosARsinA2RsinCcosB= \frac{2R\sin C\sin B - 3R\cos A}{R\sin A - 2R\sin C\cos B}

    =2sinBsinC2cos[π(B+C)]sin[π(B+C)2sinCcosB]= \frac{2\sin B\sin C - 2\cos[\pi - (B + C)]}{\sin[\pi - (B + C)- 2\sin C\cos B]}

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

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

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

  5. The diagram is given below:

    Problem 155

    Let ABCABC be the triangle. Let OO and II be the circumcenter and in-center of the ABC.\triangle ABC. Let PP be the center and x,x, the radius of the circle drawn which touches the inscribed and circumscribed circle, of ABC\triangle ABC and the side BCBC externally. Let us join OPOP and extend up to Q.Q. Let IDBC.ID\perp BC. Clearly, PP will lie on the extended part of ID.ID. Draw the line ONON parallel to the line IP.IP. Join NP.NP.

    Clearly, OB=OCOQ=R,OP=OQPQ=RxOB = OC OQ = R, OP = OQ - PQ = R - x

    ON=OM+MN=RcosA+xON = OM + MN = R\cos A + x

    NP=MD=OMCD=a2rcotC2NP = MD = OM - CD = \frac{a}{2} - r\cot \frac{C}{2}

    =a2Δs.s(sc)Δ=a2(sc)= \frac{a}{2} - \frac{\Delta}{s}.\frac{s(s - c)}{\Delta} = \frac{a}{2} - (s - c)

    =cb2= \frac{c - b}{2}

    From right angled ONP,OP2=ON2+NP2\triangle ONP, OP^2 = ON^2 + NP^2

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

    R2+x22Rx=R2cos2A+x2+2RxcosA+(cb2)2R^2 + x^2 - 2Rx = R^2\cos^2A + x^2 + 2Rx\cos A + \left(\frac{c - b}{2}\right)^2

    2Rx(1+cosA)=R2(1cos2A)(cb2)22Rx(1 + \cos A) = R^2(1 - \cos^2A) - \left(\frac{c - b}{2}\right)^2

    4Rxcos2A2=R2sin2A(bc2)2=a24(bc)244Rx\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}

    4Rxs(sa)bc=a+bc2.ab+c2=(sb)(sc)4Rx\frac{s(s - a)}{bc} = \frac{a +b - c}{2}.\frac{a - b + c}{2} = (s - b)(s - c)

    x=Δatan2A2x = \frac{\Delta}{a}\tan^2\frac{A}{2}

  6. The diagram is given below:

    Problem 156

    Since angles n the same segment of a circle are equal. BED=BAD=A2\therefore \angle BED= \angle BAD = \frac{A}{2}

    and BEF=BCF=C2\angle BEF = \angle BCF = \frac{C}{2}

    Now DEF=BEF+BED=C2+A2=C+A2=90B2\angle DEF = \angle BEF + \angle BED = \frac{C}{2} + \frac{A}{2} = \frac{C + A}{2} = 90^\circ - \frac{B}{2}

    Similarly, DFE=90C2,EDF=90A2DFE = 90^\circ - \frac{C}{2}, \angle EDF = 90^\circ - \frac{A}{2}

    Now area of DEF=12DE.DF.sinEDF\triangle DEF = \frac{1}{2}DE.DF.\sin \angle EDF

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

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

    DEsinDFE=DFsinDEF=EFsinEDF=2R\frac{DE}{\sin DFE} = \frac{DF}{\sin DEF} = \frac{EF}{\sin EDF} = 2R

    DE=2RsinDFE=2Rsin(90C2)=2RcosC2\Rightarrow DE = 2R\sin DFE = 2R\sin\left(90^\circ - \frac{C}{2}\right) = 2R\cos\frac{C}{2}

    Similarly, DF=2RcosB2DF = 2R\cos\frac{B}{2}

    So, area of DEF=12.4R2cosB2cosC2.sin(90A2)\triangle DEF = \frac{1}{2}.4R^2\cos\frac{B}{2}\cos\frac{C}{2}.\sin\left(90^\circ - \frac{A}{2}\right)

    =2R2cosA2cosB2cosC2= 2R^2\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}

    =2R2s(sa)bcs(sb)cas(sc)ab= 2R^2\sqrt{\frac{s(s - a)}{bc}}\sqrt{\frac{s(s - b)}{ca}}\sqrt{\frac{s(s - c)}{ab}}

    =2R2ss(sa)(sb)(sc)abc=2R2sΔabc=R2.s.4Δ2abc=R2s2abc4Δ= \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 ABC=Δ\triangle ABC = \Delta

    =R2s2R=R2s= \frac{R^2s}{2R} = \frac{R}{2}s

    ΔDEFΔABC=Rs2Δ=R2r\Rightarrow \frac{\Delta DEF}{\Delta ABC} = \frac{Rs}{2\Delta} = \frac{R}{2r}

  7. The diagram is given below:

    Problem 157

    ΔABC=ΔABC(ΔACB+ΔBAC+ΔCAB)\Delta A'B'C' = \Delta ABC - (\Delta AC'B' + \Delta BA'C' + \Delta CA'B')

    ΔACB=12AC.ABsinA\Delta AC'B' = \frac{1}{2}AC'.AB'\sin A

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

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

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

    AC+CB=ABc=ak+bkk=ca+bAC' + C'B = AB \Rightarrow c = ak + bk \Rightarrow k = \frac{c}{a + b}

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

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

    ΔACB=12bca+bbca+csinA\Delta AC'B' = \frac{1}{2}\frac{bc}{a + b}\frac{bc}{a + c}\sin A

    Let Δ\Delta be the area of ABC,\triangle ABC, then Δ=12bcsinA\Delta = \frac{1}{2}bc\sin A

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

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

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

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

    ΔABCΔABC=2sinAsinBsinC(sinA+sinB)(sinB+sinC)(sinC+sinA)\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)}

    =2sinA2sinB2sinC2cosAB2cosBC2cosCA2= \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:

    Problem 158

    Let the given triangle be ABCABC and the similar triangle inscribed in triangle ABCABC be ABCA'B'C' such that

    A=A,B=B,C=CA=A', B=B', C=C'

    Let BC=λa,AC=λb,AB=λcB'C'=\lambda a, A'C'=\lambda b, A'B' = \lambda c

    According to the quuestion BOC=θ\angle B'OC=\theta

    Clearly, OCB=Bθ=ACB\angle OC'B = B - \theta = \angle AC'B'

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

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

    ABC=180(C+θ+B)=Aθ\angle A'B'C' = 180^\circ - (C + \theta + B) = A - \theta

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

    BAsin(A+θ)=λbsinBBA=λbsinBsin(A+θ)\frac{BA'}{\sin(A +\theta)} = \frac{\lambda b}{\sin B}\Rightarrow BA' = \frac{\lambda b}{\sin B}\sin(A + \theta)

    BA=λ2Rsin(A+θ)\Rightarrow BA' = \lambda 2R\sin(A + \theta)

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

    ACsin(Aθ)=λcsinC\frac{A'C}{\sin(A - \theta)} = \frac{\lambda c}{\sin C}

    AC=λ2Rsin(Aθ)\Rightarrow A'C = \lambda 2R\sin(A - \theta)

    BC=BA+ACBC = BA' + A'C

    a=λ2Rsin(A+θ)+λ2Rsin(Aθ)\Rightarrow a = \lambda 2R\sin(A + \theta) + \lambda 2R\sin(A - \theta)

    a=2Rλ[sin(A+θ)+sin(Aθ)]=2Rλ2sinAcosθ\Rightarrow a = 2R\lambda[\sin(A + \theta) + \sin(A - \theta)] = 2R\lambda 2\sin A\cos\theta

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

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

  9. We have to prove that r1+r2+r3r=4Rr_1 + r_2 + r_3 - r = 4R

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

    =Δ[sa+sb(sa)(sb)+ss+cs(sc)]= \Delta\left[\frac{s - a + s - b}{(s - a)(s - b)} + \frac{s - s + c}{s(s - c)}\right]

    =Δ[c(sa)(sb)+cs(sc)]=\Delta\left[\frac{c}{(s - a)(s - b)} + \frac{c}{s(s - c)}\right]

    =Δ.c[s(sc)+(sa)(sb)s(sa)(sb)(sc)]=\Delta.c \left[\frac{s(s - c) + (s - a)(s - b)}{s(s - a)(s - b)(s - c)}\right]

    =Δ.c.1Δ2[s2sc+s2(a+b)s+ab]=\Delta.c.\frac{1}{\Delta^2}[s^2 - sc + s^2 - (a + b)s + ab]

    =cΔ[2s2s(a+b+c)+ab]=cΔ[2s22s2+ab]=\frac{c}{\Delta}[2s^2 - s(a + b + c) + ab] = \frac{c}{\Delta}[2s^2 - 2s^2 + ab]

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

  10. We have to prove that 1r1+1r2+1r3=1r\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}

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

    =3s(a+b+c)Δ=3s2sΔ[2s=a+b+c]= \frac{3s - (a + b + c)}{\Delta} = \frac{3s - 2s}{\Delta} [\because 2s = a + b + c]

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

  11. We have to prove that 1r12+1r22+1r32+1r2=a2+b2+c2Δ2\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. =1r12+1r22+1r32+1r2= \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2}

    =(sa)2+(sb)2+(sc)2+s2Δ2= \frac{(s - a)^2 + (s - b)^2 + (s - c)^2 + s^2}{\Delta^2}

    =4s22s(a+b+c)+a2+b2+c2Δ2= \frac{4s^2 - 2s(a + b + c) + a^2 + b^2 + c^2}{\Delta^2}

    =4s22s.2s+a2+b2+c2Δ2=\frac{4s^2 - 2s.2s + a^2 + b^2 + c^2}{\Delta^2}

    =a2+b2+c2Δ2== \frac{a^2 + b^2 + c^2}{\Delta^2} = R.H.S.

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

    We know that tanA2=(sb)(sc)s(sa)\tan \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} and Δ=s(sa)(sb)(sc)\Delta = \sqrt{s(s - a)(s - b)(s - c)}

    r=(sa)tanA3\Rightarrow r = (s - a)\tan\frac{A}{3}

    Similarly, r=(sb)tanB2,r=(sc)tanC2r = (s - b)\tan\frac{B}{2}, r = (s - c)\tan\frac{C}{2}

  13. We have to prove that 1A=1A1+1A2+1A3\frac{1}{\sqrt{A}} = \frac{1}{\sqrt{A_1}} + \frac{1}{\sqrt{A_2}} + \frac{1}{\sqrt{A_3}}

    R.H.S. =1A1+1A2+1A3= \frac{1}{\sqrt{A_1}} + \frac{1}{\sqrt{A_2}} + \frac{1}{\sqrt{A_3}}

    =1π(1r1+1r2+1r3)= \frac{1}{\sqrt{\pi}}\left(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\right)

    =1π(saΔ+sbΔ+scΔ)=\frac{1}{\sqrt{\pi}}\left(\frac{s - a}{\Delta} + \frac{s - b}{\Delta} + \frac{s - c}{\Delta}\right)

    =1π(3s(a+b+c)Δ)=1πsΔ=1π.1r= \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}

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

  14. r1bc+r2ca+r3ab=stanA2bc+stanB2ca=stanC2ab\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}

    =sabc(atanA2+btanB2+ctanC2)= \frac{s}{abc}\left(a\tan\frac{A}{2} + b\tan\frac{B}{2} + c\tan\frac{C}{2}\right)

    =sabc(2RsinAtanA2+2RsinBtanB2+2RsinCtanC2)= \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)

    =sabc.4R(sin2A2+sin2B2+sin2C2)= \frac{s}{abc}.4R\left(\sin^2\frac{A}{2} + \sin^2\frac{B}{2} + \sin^2\frac{C}{2}\right)

    =sΔ(1cosA2+1cosB2+1cosC2)= \frac{s}{\Delta}\left(\frac{1 - \cos A}{2} + \frac{1 - \cos B}{2} + \frac{1 - \cos C}{2}\right)

    =1r(32cosA+cosB+cosC2)= \frac{1}{r}\left(\frac{3}{2} - \frac{\cos A + \cos B + \cos C}{2}\right)

    We know that cosA+cosB+cosC=1+rR\cos A + \cos B + \cos C = 1 + \frac{r}{R}

    =1r(1r2R)=1r12R== \frac{1}{r}\left(1 - \frac{r}{2R}\right) = \frac{1}{r} - \frac{1}{2R} = R.H.S.

  15. Let DD be the point where perpendicular from AA meets BC.BC. Then AD=hAD = h

    The diagram is given below:

    Problem 165

    Clearly, OB=r,AD=h,OD=hrOB = r, AD = h, OD=h - r (If OO is below BDBD then OD=rhOD = r - h)

    BD=OB2OB2=r2(hr)2=2rhh2BD = \sqrt{OB^2 - OB^2} = \sqrt{r^2 - (h - r)^2} = \sqrt{2rh - h^2}

    Area of triangle =12.2.BD.h=h2rhh2= \frac{1}{2}.2.BD.h = h\sqrt{2rh - h^2}

  16. Let the sides be a,b,ca, b, c then Δ=12ap1=12bp2=12cp3\Delta = \frac{1}{2}ap_1 = \frac{1}{2}bp_2 = \frac{1}{2}cp_3

    p1=2Δa,p2=2Δb,p3=2Δc\Rightarrow p_1 = \frac{2\Delta}{a}, p_2 = \frac{2\Delta}{b}, p_3 = \frac{2\Delta}{c}

    L.H.S. =cosAp1+cosBp2+cosCp3= \frac{\cos A}{p_1} + \frac{\cos B}{p_2} + \frac{\cos C}{p_3}

    =12Δ[acosA+bcosB+ccosC]= \frac{1}{2\Delta}[a\cos A + b\cos B + c\cos C]

    =2R2Δ[sinAcosA+sinBcosB+sinCcosC]= \frac{2R}{2\Delta}[\sin A\cos A + \sin B\cos B + \sin C\cos C]

    =R2Δ[sin2A+sin2B+sin2C]=R2Δ4sinAsinBsinC= \frac{R}{2\Delta}[\sin 2A + \sin 2B + \sin 2C] = \frac{R}{2\Delta}4\sin A\sin B\sin C

    =abc4Δ.1R2=RR2=1R== \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. =r1r2r3=ΔsaΔsb.Δsc= r_1r_2r_3 = \frac{\Delta}{s - a}\frac{\Delta}{s - b}.\frac{\Delta}{s - c}

    =Δ3(sa)(sb))(sc)=Δ.s= \frac{\Delta^3}{(s - a)(s - b))(s - c)} = \Delta .s

    R.H.S. =r3cot2A2cot2B2cot2C2= r^3\cot^2\frac{A}{2}\cot^2\frac{B}{2}\cot^2\frac{C}{2}

    =Δ3s3.s2(sa)2Δ2.s2(sb)2Δ2.s2(sc)2Δ2= \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}

    =s3(sa)2(sb)2(sc)2Δ3=Δ.s= \frac{s^3(s - a)^2(s - b)^2(s - c)^2}{\Delta^3} = \Delta .s

  19. We have to prove that a(rr1+r2r3)=b(rr2+r3r1)=c(rr3+r1r2)=abca(rr_1 + r2r_3) = b(rr_2 + r_3r_1) = c(rr_3 + r_1r_2) = abc

    a(rr1+r2r3)=a(Δs.Δsa+Δ(sb)Δsc)a(rr_1 + r_2r_3) = a\left(\frac{\Delta}{s}.\frac{\Delta}{s - a} + \frac{\Delta}{(s - b)}\frac{\Delta}{s - c}\right)

    =Δ2.a((sb)(sc)+s(sa)s(sa)(sb)(sc))= \Delta^2.a\left(\frac{(s - b)(s - c) + s(s - a)}{s(s - a)(s - b)(s - c)}\right)

    =Δ2.a(2s2s(a+b+c)+bcΔ2)= \Delta^2.a\left(\frac{2s^2 - s(a + b + c) + bc}{\Delta^2}\right)

    =abc= abc

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

  20. We have to prove that (r1+r2)tanC2=(r3r)cotC2=c(r_1 + r_2)\tan\frac{C}{2} = (r_3 - r)\cot\frac{C}{2} = c

    (r1+r2)tanC2=(Δsa+Δsb)(sa)(sb)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}

    =sb+sa=c= s - b + s - a = c

    (r3r)cotC2=(ΔscΔs)Δ(sa)(sb)(r_3 - r)\cot\frac{C}{2} = \left(\frac{\Delta}{s - c} - \frac{\Delta}{s}\right)\frac{\Delta}{(s - a)( s - b)}

    =Δ2(ss+cs(sa)(sb)(sc))=c= \Delta^2\left(\frac{s - s + c}{s(s - a)(s - b)(s - c)}\right) = c

  21. We have to prove that 4RsinAsinBsinC=acosA+bcosB+ccosC4R\sin A\sin B\sin C = a\cos A + b\cos B + c\cos C

    R.H.S. =R(2sinAcosA+2sinBcosB+2sinCcosC)=R(sin2A+sin2B+sin2C)= R(2\sin A\cos A + 2\sin B\cos B + 2\sin C\cos C) = R(\sin 2A + \sin 2B + \sin 2C)

    =R(2sin(A+B)cos(AB)+2sinCcosC)=2R(sinCcos(AB)+sinCcosC)[sin(A+B)=sin(πC)=sinC]= 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]

    =2RsinC[cos(AB)cos(A+B)][cosC=cos(πAB)=cos(A+B)]= 2R\sin C[\cos(A - B) - \cos(A + B)][\because \cos C = \cos(\pi - A - B) = -\cos(A + B)]

    =2RsinC.2sinAsinB=4RsinAsinBsinC== 2R\sin C. 2\sin A\sin B = 4R\sin A\sin B\sin C = L.H.S.

  22. We have to prove that (r1r)(r2r)(r3r)=4Rr2(r_1 - r)(r_2 - r)(r_3 - r) = 4Rr^2

    L.H.S. =(ΔsaΔs)(ΔsbΔS)(ΔscΔs)= \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)

    =Δ3(ss+as(sa))(ss+bs(sb))(ss+cs(sc))= \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)

    =Δ3.abcs3(sa)(sb)(sc)=Δ3.abcs2Δ2=abc.Δs2= \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}

    =abcΔ.Δ2s2=4Rr2== \frac{abc}{\Delta}.\frac{\Delta^2}{s^2} = 4Rr^2 = R.H.S.

  23. We have to prove that r2+r12+r22+r32=16R2a2b2c2r^2 + r_1^2 + r_2^2 + r_3^2 = 16R^2 - a^2 - b^2 - c^2

    (r1+r2+r3r)2=r12+r22+r32+r22r(r1+r2+r3)+2(r1r2+r2r3+r3r1)(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 r1+r2+r3r=4Rr_1 + r_2 + r_3 - r = 4R and (r1r2+r2r3+r3r1)=s2(r_1r_2 + r_2r_3 + r_3r_1) = s^2

    r(r1+r2+r3)=Δs(Δsa+Δsb+Δsc)r(r_1 + r_2 + r_3) = \frac{\Delta}{s}\left(\frac{\Delta}{s -a} + \frac{\Delta}{s - b} + \frac{\Delta}{s - c}\right)

    =Δ2s(sa)+Δ2s(sb)+Δ2s(sc)=s2+(ab+bc+ca)= \frac{\Delta^2}{s(s - a)} + \frac{\Delta^2}{s(s - b)} + \frac{\Delta^2}{s(s - c)} = -s^2 + (ab + bc + ca)

    16R2=r12+r22+r32+r22[s2+(ab+bc+ca)]+2s216R^2 = r_1^2 + r_2^2 + r_3^2 + r^2 -2[-s^2 + (ab + bc + ca)] + 2s^2

    r2+r12+r22+r32=16R2a2b2c2\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=rcosecA2,IB=rcosecB2,IC=rcosecC2IA = r\cosec\frac{A}{2}, IB=r\cosec\frac{B}{2}, IC=r\cosec\frac{C}{2}

    L.H.S. =r3sinA2sinB2sinC2=r3.4R4RsinA2sinB2sinC2= \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}}

    =r3.4Rr=4R.r2=4R.Δ2s2=abcΔs2= \frac{r^3.4R}{r} = 4R.r^2 = 4R.\frac{\Delta^2}{s^2} = \frac{abc\Delta}{s^2}

    R.H.S. =abc.(sa)2(sb)2(sc)2Δ3=abcΔs2= 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 AI1=r1cosecA2AI_1 = r_1\cosec\frac{A}{2}

  26. II1=AI1AI=r1cosecA2rcosecA2II_1 = AI_1 - AI = r_1\cosec\frac{A}{2} - r\cosec\frac{A}{2}

    =(ΔsaΔs)cosecA2= \left(\frac{\Delta}{s - a} - \frac{\Delta}{s}\right)\cosec\frac{A}{2}

    =Δ.as(sa)bc(sb)(sc)=asecA3= \Delta .\frac{a}{s(s - a)}\sqrt{\frac{bc}{(s - b)(s - c)}} = a\sec\frac{A}{3}

  27. If E2E_2 be the point of contact of the circle whose center is I2I_2 with the side ACAC of the triangle ABC,ABC, we have

    AI2=AE2secI2AE2=AE2sec(90A2)=(sb)cosecA2AI_2 = AE_2\sec I_2AE_2 = AE_2sec\left(90^\circ - \frac{A}{2}\right) = (s - b)\cosec \frac{A}{2}

    I2I3=AI2+AI3=(sb+sc)cosecA2=acosecA2I_2I_3 = AI_2 + AI_3 = (s - b + s - c)\cosec\frac{A}{2} = a\cosec\frac{A}{2}