Question

In Fig. 7.58, $\Delta ABC$ is a triangle such that $\frac{AB}{AC}=\frac{BD}{DC},\angle B={70}^{\circ },\angle C={50}^{\circ }$. Find $\angle BAD$.

Answer 1

To construct a quad ABCD,

In triangle ABC where angle B =70 degree and angle C =50 degree, and

Given AB = BD and AC =DC (similar as AB/AC=BD/DC)

Therefore, Quad ABCD consists of 2 congruent scalene triangles opposite each other.

To find angle BAD, Join AD in quad ABCD. Therefore, 2 different isosceles triangles are constructed - triangle BAD and triangle CAD. In triangle ABC, angle B = 70 degree (given), angle ABD = 70 x 2 = 140 degree (property of isosceles triangle).

Therefore, angle BAD = (180 - 140) ÷ 2 degree = 20 degree.

Metacognitive analysis:

To prove that angle BAD= 20 degree

Similarly, angle CAD = 40 degree.

Sum of angles in a QUAD = 360 degree :

Angle A = 60 + Angle B 140 + Angle C 100 + Angle D =60

Therefore, angle BAD is 20 degree.

Answer: 20 degree