In figure BD and CE intersect each other at the point P. Is`DeltaPBC~DeltaPDE`? Why?

Explanation

True
In `DeltaPBC and DeltaPDE`,
`angleBPC=angleEPD` [vertically opposite angle]
Now `(PB)/(PD)=5/10=1/2`
and `(PC)/(PE)=(6)/(12)=1/2`
From Eqs. (i) and (ii) `(PB)/(PD)=(PC)/(PE)`
Since. One angle of `DeltaPBC` is equal to one angle of `DeltaPDE` and the sides including these angles are proportional, so both triangles are similar.
Hence. `DeltaPBC~DeltaPDE`, by SAS similarity criterion.