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# ABCD is a rhombus such that the perpendicular bisector of AB passes through D. Find the angles of the rhombus.

Hint: Join BD. Then ∆ ABD is equilateral.

**Solution:**

Given, ABCD is a __rhombus__.

The __perpendicular bisector__ of AB passes through D.

We have to find the angles of the rhombus.

DE is the altitude on AB.

Now, AE = EB.

Considering ΔAED and ΔBED,

Common side = DE

∠AED = ∠BED = 90°

Given, AE = EB

By __SAS__ criterion, ΔAED ≅ ΔBED

By CPCT,

AD = BD ------------------- (1)

We know that all sides of a rhombus are equal.

So, AD = AB ------------- (2)

From (1) and (2),

AD = AB = BD

Since the three sides are equal, ABD is an equilateral triangle.

In an __equilateral triangle__, each angle is equal to 60 degrees.

So, ∠A = 60°

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

So, ∠A = ∠C = 60°

We know that the sum of adjacent angles of a rhombus is __supplementary__.

So, ∠ABC + ∠BCD = 180°

∠ABC + 60° = 180°

∠ABC = 120°

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

So, ∠ABC = ∠ADC = 120°

Therefore, the angles of rhombus are ∠A = ∠C = 60° and ∠B = ∠D = 120°.

**✦ Try This:** ABCD is parallelogram whose diagonals intersect at O. If OD = 9 cm,OB = x + y cm, OA = x - y, and OC = 5 cm, find x and y.

**☛ Also Check: **NCERT Solutions for Class 8 Maths

**NCERT Exemplar Class 8 Maths Chapter 5 Problem 179**

## ABCD is a rhombus such that the perpendicular bisector of AB passes through D. Find the angles of the rhombus. Hint: Join BD. Then ∆ ABD is equilateral

**Summary:**

ABCD is a rhombus such that the perpendicular bisector of AB passes through D. The angles of the rhombus are 60°, 120°, 60° and 120°.

**☛ Related Questions:**

- ABCD is a parallelogram. Points P and Q are taken on the sides AB and AD respectively and the parall . . . .
- In parallelogram ABCD, the angle bisector of ∠A bisects BC. Will angle bisector of B also bisect AD? . . . .
- A regular pentagon ABCDE and a square ABFG are formed on opposite sides of AB. Find ∠BCF

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