Question

If a diagonal of a parallelogram bisects one of the angles of the parallelogram then the angle
between the diagonals is​

Answer 1

Answer:

The angle between the diagonals 90°.

Step-by-step explanation:

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°.