In parallelogram abcd the bisectors of adjacent angles a and b intersect each other at point p prove that angle apd=90 degree
Answers
Given : In a parallelogram ABCD, the bisectors of adjacent angles A and D intersect each other at point P.
To Find : Prove that ∠APD = 90°
Solution:
Opposite angles of a parallelogram are equal
Sum of adjacent angles of a parallelogram = 180°
=> ∠A + ∠D = 180°
=> (1/2)(∠A +∠D) = 90°
AP is angle bisector of ∠A
∠DAP =∠A/2
DP is angle bisector of ∠D
∠ADP =∠D/2
∠DAP + ∠ADP + ∠APD = 180° ( sum of angles of a triangle )
=> ∠A/2 + ∠D/2 + ∠APD = 180°
=> (1/2)(∠A +∠D) + ∠APD = 180°
=> 90° + ∠APD = 180°
=> ∠APD = 90°
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Answer:
Given : In a parallelogram ABCD, the bisectors of adjacent angles A and D intersect each other at point P.
To Find : Prove that ∠APD = 90°
Solution:
Opposite angles of a parallelogram are equal
Sum of adjacent angles of a parallelogram = 180°
=> ∠A + ∠D = 180°
=> (1/2)(∠A +∠D) = 90°
AP is angle bisector of ∠A
∠DAP =∠A/2
DP is angle bisector of ∠D
∠ADP =∠D/2
∠DAP + ∠ADP + ∠APD = 180° ( sum of angles of a triangle )
=> ∠A/2 + ∠D/2 + ∠APD = 180°
=> (1/2)(∠A +∠D) + ∠APD = 180°
=> 90° + ∠APD = 180°
=> ∠APD = 90°