Question

Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angkle QPR meets QR at point M, hence, $\frac{AB}{PQ}=\frac{AD}{PM}$.
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

Given: $△ABC\sim △PQR$
$\angle ABC=\angle PQR$ (Corresponding angles) (I)
$\angle BAC=\angle QPR$ (Corresponding angles)
$\frac{1}{2}\angle BAC=\frac{1}{2}\angle QPR$ (II)
Now, AD and PM are angle bisectores of $\angle BAC$ and $\angle QPR$ respectively,
thus, in $△ABD$ and $△PQM$
$\angle ABD=\angle PQM$ (From I)
$\angle BAD=\angle QPM$ (From II)
$\angle BDA=\angle QMP$ (Third angle)
Thus, $△ABD\sim △PQM$ (AAA rule)
Hence, $\frac{AB}{PQ}=\frac{AD}{PM}$