Question

Question 4
In the figure, BD and CE intersect each other at the point P. Is $\Delta PBC\sim \Delta PDE$? Why?

Answer 1

In $\Delta PBCand\Delta PDE$,
$\angle BPC=\angle EPD$ [Vertically opposite angles ]
Now, $\frac{PB}{PD}=\frac{5}{10}=\frac{1}{2}$ ……(i)
and $\frac{PC}{PE}=\frac{6}{12}=\frac{1}{2}$ …..(ii)
From eqs. (i) and (ii) $\frac{PB}{PD}=\frac{PC}{PE}$
Since, one angle of $\Delta PBC$ is equal to one angle of $\Delta PDE$ and the sides including the angle are proportional, both the triangles are similar.
Hence, $\Delta PBC\sim \Delta PDE$ by SAS similarity criterion.