Given P is the midpoint of AE P is the midpoint of BK ABEK A

Given: P is the midpoint of AE P is the midpoint of BK AB||EK AB EK Prove: delta ABP = delta EKP

Solution

Given P is the midpoint of AE,
Therefore AP = PE.

Given P is the midpoint of BK,
Therefore BP = PK.

Given, AB is parallel to EK and AB = EK.
m(BPA) = m(EPK) as they are vertically opposite angles.

In triangles APB and EPK,
AP = EP,
m(BPA) = m(EPK),
BP = KP
Therefore, by SAS congruency property both the traingles are congruent.