Question

In the adjoining figure, ABC and DBC are two isosceles triangles on the same base BC. Prove that ∠ABD = ∠ACD.

Answer 1

Given: ABC and DBC are two isosceles triangles on the same base BC, so AB = AC and DB = DC.

To Prove: ∠ABD = ∠ACD.

Proof [Step-wise]:

1. In isosceles triangle ABC, the base angles are equal; hence ∠ABC = ∠BCA.

2. In isosceles triangle DBC, the base angles are equal; hence ∠CBD = ∠BCD.

3. At vertex B the angle between BA and BD (∠ABD) is the sum of the angle between BA and BC and the angle between BC and BD.

Thus, ∠ABD = ∠ABC + ∠CBD.

4. At vertex C the angle between CA and CD (∠ACD) is the sum of the angle between CA and CB and the angle between CB and CD.

Thus, ∠ACD = ∠BCA + ∠BCD.

5. Substitute equalities from steps 1 and 2 into steps 3 and 4:

∠ABD = ∠ABC + ∠CBD

= ∠BCA + ∠BCD

= ∠ACD

Therefore ∠ABD = ∠ACD, as required.