Question

In the given figure, $2AP=5PB$ and $2CP=5PD$

Then, $\Delta ACP$ and $\Delta BDP$ are similar.
If the above statement is true then mention answer as 1, else mention 0 if false.

Answer 1

Given, 2 PA = 5 PB and 2 CP = 5 PD

hence, $\frac{PA}{PB}=\frac{5}{2}$ and $\frac{PC}{PD}=\frac{5}{2}$

Now, in $△$PAC and PDB,

$\angle APC=\angle BPD$ (Vertically opposite angles)

$\frac{PA}{PB}=\frac{PC}{PD}$

Thus, $△APC\sim △BPD$ (SAS rule)