Question

Triangle ABC is similar to triangle PQR. The bisector of angle BAC meets BC at point D and the bisector of angle QPR meets QR at point M. Which of the following is correct?

• A. $\frac{AB}{PQ}=\frac{AD}{PM}$
• B. $\frac{PQ}{AB}=\frac{AD}{PM}$
• C. $\frac{AB}{AD}=\frac{PQ}{PM}$
• D. $\frac{AB}{AD}=\frac{PM}{PQ}$

Answer 1

Answer: (A), (C)

The correct options are
A

$\frac{AB}{PQ}=\frac{AD}{PM}$

C

$\frac{AB}{AD}=\frac{PQ}{PM}$

Given, $\Delta ABC\sim \Delta PQR$ and AD and PM are the angle bisectors of $\angle A$ and $\angle P$ respectively.

Since $\Delta ABC\sim \Delta PQR$, we have,

$\angle A=\angle P$

$\angle B=\angle Q$

$\frac{AB}{PQ}=\frac{BC}{QR}$

$\because \angle A=\angle P\Rightarrow \frac{1}{2}\angle A=\frac{1}{2}\angle P$

$\Rightarrow \angle BAD=\angle QPM$

Now in $\Delta ABD$ and $\Delta PQM$

$\angle B=\angle Q$

$\angle BAD=\angle QPM$ (Proved)

$\therefore \Delta BAD\sim \Delta QPM$ (AA axiom)

$\therefore \frac{AB}{PQ}=\frac{AD}{PM}$