Question

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

Answer 1

Given:

ABCD is a parallelogram.

Diagonal AC bisects ∠A and also bisects ∠C.

To Prove:

The two diagonals AC and BD are perpendicular to AC ⟂ BD.

Proof (Step-wise):

1. Let AC meet BD at O.   ...(O is the intersection of the diagonals)

2. Since AC bisects ∠A, we have ∠BAC = ∠CAD.

Since AC bisects ∠C, we have ∠BCA = ∠DCA.

3. In triangle ABC,

∠BAC = ∠BCA   ...(From step 2) 

Hence, triangle ABC is isosceles, so AB = BC.

4. In a parallelogram, opposite sides are equal.

So, BC = AD and AB = CD. 

From AB = BC and BC = AD we get AB = AD.

Thus, AB = BC = CD = DA; all four sides are equal ABCD is a rhombus.

5. In any parallelogram, diagonals bisect each other.

So, AO = OC and BO = OD.

6. Consider triangles AOB and COB:

AO = CO   ...(Step 5)

AB = CB   ...(Step 3)

OB = OB   ...(Common)

Therefore, triangles AOB and COB are congruent by SSS.

7. From the congruence,

∠AOB = ∠BOC

But ∠AOB and ∠BOC are adjacent angles that sum to 180°.

So, each equals 90°.

8. Hence, ∠AOB = 90°.

So, AC ⟂ BD.

Therefore, if a diagonal of a parallelogram bisects one angle and also bisects the opposite angle, the diagonals of the parallelogram are perpendicular to each other.