Then because △ABE is an isosceles triangle, ∠ e = ∠ BAE =1AEC/2 ∠ Abd = ∠ C.

In △AED and △ACD, ∠E=∠C, ∠ ADE = ∠ ADC = 90, and AD is the common side.

So △ aed △ ACD

So CD=ED=BD+BE=BD+AB.

2. Prove that in equilateral △ABC,

AB=AC，∠BAC =∠ACB=60，

Namely: ∠BAD=∠ACE

And \ad = CE in △BAD and △ACE,

∴△bad?△ace(SAS)

∴ ∠ABD=∠CAE，

Namely: ∠ABP=∠CAE

∴∠bpq =∠ABP+∠BAP =∠CAE+∠BAE =∠BAC = 60

And BQ⊥AE

∴ ∠BQP=90

∴ ∠PBQ=30

∴ BP=2PQ。

That is, PO= 1/2BP.