Question

State true or false:

Triangle $ABC$ is similar to triangle $PQR$. If bisector of $\angle BAC$ meets $BC$ at point $D$ and the bisector of $\angle QPR$ meets $QR$ at point $M$, Then, $\frac{AB}{PQ}=\frac{AC}{PM}.$

• A. True
• B. False

Answer 1

The correct option is B False

Given: $△ABC\sim △PQR$
$\angle ABC=\angle PQR$ (Corresponding angles) (I)
$\angle BAC=\angle QPR$ (Corresponding angles)
$\Rightarrow \frac{1}{2}\angle BAC=\frac{1}{2}\angle QPR$ (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, $\frac{AB}{PQ}=\frac{AD}{PM}$