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# 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

**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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