Question

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

Options

•  70°

• 90°

• 80°

• 100°

Answer 1

Figure can be drawn as follows:

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

Also, it is given that ∠ABD = 50°

We need to find ∠DPC

It is given that∠ABD = 50°

Therefore, ∠ABP = 50°(Because P lies on BD)

Also, diagonals of rectangle are equal and they bisect each other.

Therefore,

AP = PB

Thus,∠ABP = ∠BAP (Angles opposite to equal sides are equal)

∠BAP = 50°  [∠ABP = 50° (Given)]

By angle sum property of a triangle:

∠BAP + ∠ABP + ∠APB = 180°

         50° + 50 +∠APB = 180°

              100° + ∠APB = 180°

                          ∠APB = 80°

But ∠APB and ∠DPC are vertically opposite angles.

Therefore, we get:

∠DPC = ∠APB

 ∠DPC = 80°  [(∠APB = 80° lready proved)]

Hence the correct choice is (c).