(i) ∆ABE ≅ ∆ACF
(ii) AB = AC, i.e., ∆ABC is an isosceles triangle.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∆BCD is a right angle.
Given: An equilateral triangle ABC.
To Prove: ∠A = ∠B = ∠C = 60°.
Proof: ∵ ABC is an equilateral triangle
∴ AB = BC = CA ...(1)
∵ AB = BC
∴ ∠A = ∠C ...(2)
| Angles opposite to equal sides of a triangle are equal
∵ BC = CA
∴ ∠A = ∠B ...(3)
| Angles opposite to equal sides of a triangle are equal
From (2) and (3), we obtain
∠A = ∠B = ∠C ...(4)
In ∆ABC,
∠A + ∠B + ∠C = 180° ...(5)
| Sum of all the angles of a triangle is 180°
Let ∠A = x°. Then, ∠B = ∠C = x°
| From (4)
From (5),
x° + x° + x° = 180°
3x° = 180°
⇒ x° = 60°
⇒ ∠A = ∠B = ∠C = 60°.
ABC is a triangle in which ∠B = 2∠C. D is a point on side BC such that AD bisects ∠BAC and AB = CD. Prove that ∠BAC = 72°.
[Hint. Take a point P on AC such that BP bisects ∠B. Join P and D.]