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$ABC$ is a cyclic triangle and bisector of angle $B\widehat{A}C$, $A\widehat{B}C$ and $A\widehat{C}B$ touches circle at $P$, $Q$ and $R$ respectively then measure of angle $R\widehat{Q}P$ is?

The options are:

  1. $90-\frac{B\widehat{A}C}{2}$
  2. $90-\frac{A\widehat{B}C}{2}$
  3. $90+\frac{A\widehat{B}C}{2}$
  4. $90+\frac{A\widehat{C}B}{2}$

I know how to draw the figure for this question but I can't establish the relation between the angles of the two triangles...

AugSB
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2 Answers2

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The tool you need is an Inscribed angle theorem.

Roman83 shows how it applies to your problem.

CiaPan
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enter image description here

$\angle PQR=\angle PQB+\angle BQR=\angle BAP+\angle BCR=\frac A2+\frac C2=\frac{180-B}2=90-\frac B2$

Roman83
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