The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD =50°, then ∠DPC =
Given that,
∠ABD = ∠ABP = 50°
∠PBC +∠ABP = 90° (Each angle of a rectangle is a right angle)
∠PBC = 40°
Now,
PB = PC (Diagonals of a rectangle are equal and bisect each other)
Therefore,
∠BCP = 40° (Equal sides has equal angle)
In triangle BPC,
∠BPC + ∠PBC + ∠BCP = 180° (Angle sum property of a triangle)
∠BPC = 100°
∠BPC + ∠DPC = 180° (Angles in a straight line)
∠DPC = 180o – 100o
= 80°