Question

In the adjoining figure, ABCD is a rhombus and ABE is an equilateral triangle. ∠BCD = 70°, find:

1. ∠ADE
2. ∠BDE
3. ∠BED

Answer 1

Given:

ABCD is a rhombus, ABE is an equilateral triangle and ∠BCD = 70°.

Step-wise calculation:

1. From the rhombus:

Opposite angles are equal and adjacent angles are supplementary. 

So, ∠A = ∠C = 70° and ∠B = ∠D = 110°.

Also, a diagonal of a rhombus bisects the vertex angles.

So, diagonal DB bisects ∠D, giving `∠BDA = 1/2 xx110^circ = 55^circ`.

2. From the equilateral triangle ABE: 

All sides AB = BE = AE and all angles at A, B, E are 60°.

In particular, AE = AB.   ...(Equilateral-triangle)

3. Because ABCD is a rhombus AB = AD and from step 2 AE = AB, we have AD = AE.

Thus, triangle ADE is isosceles with AD = AE.

So, its base angles at D and E are equal.

4. The angle at A of triangle ADE is the angle between AD and AE.

At A, we pass from AD to AB (70° inside the rhombus), then from AB to AE (60° in the equilateral). 

So, ∠DAE 

= 70° + 60° 

= 130°

Therefore, in triangle ADE:

130° + 2 × ∠ADE = 180°

⇒ 2 × ∠ADE = 50°

⇒ ∠ADE = 25°

5. At D, the diagonal DB makes angle ∠BDA = 55° with DA from step 1.

The angle ∠BDE between DB and DE equals ∠BDA – ∠ADE = 55° – 25° = 30°.

Since DE lies on the other side of DA than DB in the figure.

6. Finally, in triangle BDE the three angles must sum to 180°:

∠BDE + ∠DBE + ∠BED = 180°

We already know ∠DBE = 115°

This follows from the bisected angle at B (55°) combined with ∠ABE = 60° and ∠BDE = 30°. 

So, ∠BED

= 180° – 30° – 115° 

= 35°

1. ∠ADE = 25° 
2. ∠BDE = 30° 
3. ∠BED = 35°