The diagonals AC and BD of a rectangle ABCD intersect each other at P.

1.	The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, then ∠DPC = A. 70° B. 90° C. 80° D. 100°
Answer» Option : (C) 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 = 180° – 100° = 80°

The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, then ∠DPC = A. 70° B. 90° C. 80° D. 100°

Answer»

Option : (C)

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 = 180° – 100°

= 80°