Question

In the adjoining figure, AC = BC = CD and ∠ADC = 40°. Find ∠BAC.

Answer 1

Given: AC = BC = CD, points B, C, D are collinear and ∠ADC = 40°.

Step-wise calculation:

1. From AC = BC,

Triangle ABC is isosceles with AC = BC,

So, ∠CAB = ∠ABC. 

Let ∠BAC = x,

Hence, ∠ABC = x.

2. From AC = CD,

Triangle ACD is isosceles with AC = CD, 

So, base angles at A and D are equal: ∠CAD = ∠ADC.

3. Given ∠ADC = 40°,

Therefore, ∠CAD = 40°.

4. At A the full angle between AB and AD is ∠BAD

= ∠BAC + ∠CAD

= x + 40°

5. In triangle ABD,

The three angles sum to 180°. 

The angle at B is ∠ABD = ∠ABC = x since B–C–D are collinear, ray BD is the same base direction as BC. 

The angle at D in triangle ABD is ∠ADB, which equals ∠ADC because DC and DB are collinear.

So, ∠ADB = 40°.

6. Sum for triangle ABD:

(Angle at A) + (Angle at B) + (Angle at D)

= (x + 40°) + x + 40°

= 180°

7. Solve:

2x + 80° = 180°

⇒ 2x = 100°

⇒ x = 50°

⇒ ∠BAC = 50°