Question

In figure, BD and CE intersect each other at the point P. Is ΔPBC ~ ΔPDE? Why?

Answer 1

In ∆PBC and ∆PDE,

∠BPC = ∠EPD ...[Vertically opposite angles]

Now, `("PB")/("PD") = 5/10 = 1/2`   ...(i)

And `("PC")/("PE") = 6/12 = 1/2`   ...(ii)

From equations (i) and (ii),

`("PB")/("PD") = ("PC")/("PE")`

Since, one angle of ∆PBC is equal to one angle of ∆PDE and the sides including these angles are proportional, so both triangles are similar.

Hence, ∆PBC ~ ∆PDE, by SAS similarity criterion.