Question

In quadrilateral ABCD, seg AD || seg BC. Diagonal intersect at point P. If AP = 1/3 AC. Prove = DP = 1/2 BP

Answer 1

We have two pairs of alternate angles, which are congruent:

∠ PDA = ∠ PBC

∠ PAD = ∠ PCB

Also ∠ APD = ∠ BPD  ( vertical angles )

Therefore Δ APD ≅ Δ BPC

Those triangles are congruent.

From AP = 1/3 AC we have:

AP : PC = 1 : 2

or: AP = 1/2 PC

Therefore DP : BP = 1 : 2   or   DP = 1/2 BP

Hence proved.