Question

If AD and PM bisects $\angle BAC$ and $\angle QPR$ of $\Delta ABC$ and $\Delta PQR$ respectively and $\Delta ABC\sim \Delta PQR$ then prove that $\frac{AB}{PQ}=\frac{AD}{PM}$

Answer 1

Given $\Delta$ABC and $\Delta$PQR AD is median of $\Delta$AB & PM is median of $\Delta$ PQR & $\Delta$ ABC $\sim \Delta$PQR

Since AD is the median

BD$=$CD$=\frac{1}{2}$BC

Similarly PM is the median

QM$=$RM$=\frac{1}{2}$QR

Now $\Delta$ABC $\sim \Delta$PQR

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

So, $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{2BD}{2QM}=\frac{BD}{QM}$

also since $\Delta$ABC $\sim \Delta$ PQR

$\angle B=\angle Q$(corresponding angles of similar $\frac{AB}{PQ}=\frac{BD}{QM}$ triangles are equal)

Hence by SAS similarly

$\Delta$ABD $\sim \Delta$ PQM

Since corresponding sides of similar triangles are proportional

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