# the diagonals AC and BD of a rectangle ABCD intersect each other at P. if angle ABD = 50 degree, then angle DPC =

Given: Angle ABD = Angle ABP = 500

Angle PBC + Angle ABP = 900  (Each angle of a rectangle is a right angle)

Angle PBC = 400

Now, PB = PC  (Diagonals of a rectangle are equal and bisect each other)

Therefore,

Angle BCP = 400  (Equal sides has equal angles)

In triangle BPC,

Angle BPC + Angle PBC + Angle BCP = 1800  (Angle sum property of a triangle)

Angle BPC = 1000

Angle BPC + Angle DPC = 1800  (Angles in a straight line)

Angle DPC = 1800 - 1000 = 800.

• 22

first prove that abcd is a llgm. by the prop. that 2 pairs of opp. sides are equal then its a llgm.

now we know that the diadonals of a llgm. bisect each other.

therefore DP=PB

hence DPA+BPA=180 linear pair

DPA+DPA=180 since Dp=Bp

Dpa=180/2

DPA=90

Bpa = 90

now in triangle apb and dpc

BPA=DPC vert. opp. angles

90=DPC

• -13

the answer in this question is 80 degree.

• -10
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