Question

In adjoining fig, in parallelogram abcd, AP and DP are bisectors of angle A and angle D respectively find angle APD

Answer 1

Since quadrilateral ABCD is a parallelogram therefore
Angle BAD +angle CDA=180°.(1) (Sum of adjacent angles of a parallelogram is 180°)
Duviding (1) by 2
angle BAD/2+ angle CDA/2=180°/2 (2)

Since AP and DP are bisectors of BAD and CDA therefore
Angle PAB= angle PAD=1/2 angle BAD. And
Angle PDC=angle PDA=1/2angle CDA
Therefore, on putting this values in (2) we get
Angle PAD +Angle PDA=90° (3)
In triangle PAD
Angle PAD+ angle PDA +angle APD=180°. (Angle sum property of triangle
90°+angle APD= 180°. From (3)
Angle APD=90°