If a diagonal of a parallelogram bisects one of the angles of the parallelogram, it also bisects the second angle and then the two diagonals are perpendicular to each other.
Given: A parallelogram ABCD, in which ∠1 = ∠2 and diagonals AC and DB intersect at P.
To Prove: ∠3 = ∠4 and ∠APB = 90°
Proof: Since AD ïï BC and DB is the transversal
... ∠2 = ∠4 (alternate interior angles )......(i)
Similarly,
∠1 = ∠3 ..... (ii)
But ∠1 = ∠2 (given) ...(iii)
From equations (i), (ii) and (iii), we get,
∠3 = ∠4
Thus diagonal DB bisects ∠B also.
Now in ΔABP and ΔADP, ∠2 = ∠3
AB = AD..... (sides opposite to equal angles)
DP = PB .... (diagonals bisects each other)
and AP = AP ... (common)
ΔABP ≅ ΔADP
... ∠APD = ∠APB ...... (c.p.c.t.)
But ∠APD + ∠APB = 180° ......(linear pair)
2 ∠APB = 180° ⇒∠APB = 90°
∴ Diagonals are bisect each other at 90°.
To Prove: ∠3 = ∠4 and ∠APB = 90°
Proof: Since AD ïï BC and DB is the transversal
... ∠2 = ∠4 (alternate interior angles )......(i)
Similarly,
∠1 = ∠3 ..... (ii)
But ∠1 = ∠2 (given) ...(iii)
From equations (i), (ii) and (iii), we get,
∠3 = ∠4
Thus diagonal DB bisects ∠B also.
Now in ΔABP and ΔADP, ∠2 = ∠3
AB = AD..... (sides opposite to equal angles)
DP = PB .... (diagonals bisects each other)
and AP = AP ... (common)
ΔABP ≅ ΔADP
... ∠APD = ∠APB ...... (c.p.c.t.)
But ∠APD + ∠APB = 180° ......(linear pair)
2 ∠APB = 180° ⇒∠APB = 90°
∴ Diagonals are bisect each other at 90°.
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