A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.

Solution:

Consider a parallelogram ABCD

Join the diagonal AC

AC bisects angle A

We have to prove that ABCD is a rhombus.

Since AC bisects angle A

∠CAB = ∠CAD ------------------ (1)

We know that the opposite sides of a parallelogram are parallel and congruent.

So, AB||CD and AC is a transversal.

We know that the alternate interior angles are equal.

∠CAB = ∠ACD ----------------- (2)

Similarly, AD||BC and AC is a transversal.

∠DAC = ∠ACB ----------------- (3)

From (1), (2) and (3),

∠BCA = ∠DCA

We know that the opposite angles of a parallelogram are equal.

∠A = ∠C

Dividing by 2 on both sides,

1/2 ∠A = 1/2 ∠C

Comparing (1) and (2),

∠CAD = ∠ACD

We know that the sides opposite to the equal angles are equal.

CD = AD

We know that the opposite sides of a parallelogram are parallel and congruent.

AB = CD

AD = BC

So, AB = BC = CD = AD

This implies all the sides are equal.

Therefore, ABCD is a rhombus.

✦ Try This: ABCD is a parallelogram in which P and Q are mid-points of opposite sides AB and CD. If AQ intersects DP at S and BQ intersects CP at R, show that DPBQ is a parallelogram

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 8

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NCERT Exemplar Class 9 Maths Exercise 8.4 Problem 6

A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.

Summary:

The (interior) bisector of an angle, also called the internal angle bisector, is the line or line segment that divides the angle into two equal parts. A diagonal of a parallelogram bisects one of its angles. It is shown that it is a rhombus

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☛ Related Questions:

• P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. AQ intersects DP . . . .
• ABCD is a quadrilateral in which AB || DC and AD = BC. Prove that ∠A = ∠B and ∠C = ∠D
• In Fig. 8.11, AB || DE, AB = DE, AC || DF and AC = DF. Prove that BC || EF and BC = EF