(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.
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.]
Given: D and E are points on the base BC of a ∆ABC such that AD = AE and ∠BAD = ∠CAE.
To Prove: AB = AC
Proof: In ∆ADE,
∵ AD = AE | Given
∴ ∠ADE = ∠AED ...(1)
| Angles opposite to equal sides of a triangle are equal
In ∆ABD,
Ext. ∠ADE = ∠BAD + ∠ABD ...(2)
| An exterior angle of a triangle is equal to the sum of its two interior opposite angles
In ∆AEC,
Ext. ∠AED = ∠CAE + ∠ACE .. .(3)
| An exterior angle of a triangle is equal to the sum of its two interior opposite angles
From (1), (2) and (3),
∠BAD + ∠ABD = ∠CAE + ∠ACE
⇒ ∠ABD = ∠ACE
| ∵ ∠BAD = ∠CAE (Given)
⇒ ∠ABC = ∠ACB
∴ AB = AC
| Sides opposite to equal angles of a triangle are equal