Question

If the diagonal of a parallelogram bisects one of the angles of a parallelogram, it also bisects the second angle and then the two diagonals are perpendicular to each other.

Answer 1

Given: A ||gm ABCD whose diagonals AC bisects ∠A. Diagonals AC and BD intersectat O.

To prove: (i) ∠3 =∠4

(ii)AC and BD are perpendicular to each other,

Proof:

(i) ∠1 = ∠3 (Alternate angles, AD|| BC) ..........(i)

∠2 = ∠4 (Alternate angles, AB|| DC).......(ii)

But, ∠1=∠2

∴ ∠3=∠4

Also, ∠A=∠C (Opposite angles of a ||gm)

∴ 1/2∠A = 1/2 ∠C

∴ ∠1=∠4

In ΔACD,

∠1=∠4 (Proved above)

But, AD =CD (opposite sides of equilateral triangles)

∴ AB =BC  and CD = AB. (opposite sides of a ||gm)

∴ AB=BC = CD = AD

∴ ABCD is a rhombus.

Since, the diagonals of a rhombus are perpendicular to each other, therefore, AC and BD are perpendicular to each other.

Answer 2

Answer:

check the attachment (◍•ᴗ•◍)