The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, find ∠DPC.

The diagonals of a rectangle are equal and bisect each other.

∴ AP = BP

From figure,

In △ABP,

AP = BP

⇒ ∠PAB = ∠ABP = 50° (Angles opposite to equal sides are equal in isosceles triangle)

⇒ ∠PAB + ∠ABP + ∠APB = 180°

⇒ 50° + 50° + ∠APB = 180°

⇒ ∠APB = 180° - 100° = 80°.

⇒ ∠DPC = ∠APB = 80° (Vertically opposite angles are equal).

Hence, ∠DPC = 80°.