Math, asked by Shweta1505, 10 months ago

The triangle ABC is inscribed in a circle, the bisectors AX, BY and CZ of the angles
angleBAC, angleABC and angleACB intersect at the points X, Y, Z, on the circle respectively, let us
prove that AX is perpendicular to YZ.​

Answers

Answered by suriyasankar1974
0

Answer:

Given\quad \triangle ABC\quad is\quad inscribed\quad in\quad C(0,r).Given△ABCisinscribedinC(0,r).

The\quad bisectors\quad of\quad \angle BAC,\angle ABC\quad and\quad \angle ACB\quad meets\quad the\quad circumcircleThebisectorsof∠BAC,∠ABCand∠ACBmeetsthecircumcircle

of\quad \triangle ABC,in\quad P,Q,R\quad respectively.of△ABC,inP,Q,Rrespectively.

In\quad the\quad figure\quad Join\quad RQ,InthefigureJoinRQ,

\angle ABQ=\angle APQ\quad -(i)∠ABQ=∠APQ−(i)

\{ Angles\quad in\quad the\quad same\quad segment\quad of\quad a\quad circle\quad are\quad equal\} .{Anglesinthesamesegmentofacircleareequal}.

\angle ABQ=\angle QBC\quad \{ BQ\quad is\quad the\quad bisector\quad of\quad \angle ABC\} .∠ABQ=∠QBC{BQisthebisectorof∠ABC}.

\therefore \angle QBC=\angle APQ\quad -(ii)∴∠QBC=∠APQ−(ii)

Adding\quad (i)\& (ii)Adding(i)&(ii)

\angle ABQ+\angle QBC=\angle APQ+\angle APQ∠ABQ+∠QBC=∠APQ+∠APQ

\therefore \angle ABC=2\angle APQ\quad -(iii)∴∠ABC=2∠APQ−(iii)

Similarily,\angle ACB=2\angle APR\quad -(iv)Similarily,∠ACB=2∠APR−(iv)

Adding\quad (iii)\& (iv)Adding(iii)&(iv)

\angle ABC+\angle ACB=2(\angle APQ+\angle APR)∠ABC+∠ACB=2(∠APQ+∠APR)

\therefore \angle ABC+\angle ACB=2\angle QPR\quad -(v)∴∠ABC+∠ACB=2∠QPR−(v)

In\triangle ABC,In△ABC,

\angle ABC+\angle BAC+\angle ACB={ 180 }^{ \circ  }\quad \{ Angle\quad sum\quad property\}∠ABC+∠BAC+∠ACB=180  

∘  

{Anglesumproperty}

\angle ABC+\angle ACB={ 180 }^{ \circ  }-\angle BAC\quad -(vi)∠ABC+∠ACB=180  

∘  

−∠BAC−(vi)

From\quad (v)\& (vi),we\quad getFrom(v)&(vi),weget

{ 180 }^{ \circ  }-\angle BAC=2\angle QPR180  

∘  

−∠BAC=2∠QPR

\therefore \angle QPR=\cfrac { 1 }{ 2 } (180-\angle BAC)∴∠QPR=  

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1

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(180−∠BAC)

\angle QPR=\cfrac { 1 }{ 2 } \times { 180 }^{ \circ  }-\cfrac { 1 }{ 2 } \angle BAC∠QPR=  

2

1

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×180  

∘  

−  

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1

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∠BAC

\implies\quad \angle QPR=90-\cfrac { 1 }{ 2 } \angle BAC⟹∠QPR=90−  

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1

​  

∠BAC

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