Solution:

True

In triangles PBC and PDE,

∠EPD = ∠BPC  [ as vertically opposite angles]

$PB/PD\text{ }=\text{ }5/10\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}$… (i)

$PC/PE\text{ }=\text{ }6/12\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}$… (ii)

Now from eq.(i) and eq.(ii),

We obtain,

PC/PE = PB/PD

As it is known to us that, ∠BPC of ∆PBC = ∠EPD of ∆PDE and the sides including these.

Then, by the SAS similarity criteria

∆PBC ∼ ∆PDE