Question

In ΔABC, AB = AC, D is a point inside the triangle such that ∠DBC = ∠DCB.

Prove that ΔBAD ≅ ΔCAD.

Answer 1

Given:

In ΔABC, AB = AC (i.e., ΔABC is isosceles with AB = AC)

D is a point inside the triangle such that ∠DBC = ∠DCB

To Prove:

ΔBAD ≅ ΔCAD

Proof:

1. Since AB = AC (given), we have two equal sides in ΔABC.

2. D is inside the triangle, and ∠DBC = ∠DCB is given. This means that in triangle BDC, angles at B and C are equal.

3. Hence, in ΔBDC, the two angles ∠DBC and ∠DCB are equal, so BD = CD by the Isosceles Triangle Property.

4. Consider triangles BAD and CAD.

We know:

AB = AC   ...(Given)

AD = AD   ...(Common side)

BD = CD   ...(From Step 3)

6. Also, since ∠DBC = ∠DCB and BD = CD, angles ∠ABD and ∠ACD are equal because:

∠ABD = ∠DBC   ...(As part of the larger angle at B)

∠ACD = ∠DCB   ...(As part of the larger angle at C)

But to prove congruency, it’s more straightforward to focus on the sides and included angles.

7. In triangles BAD and CAD:

AB = AC   ...(Given)

AD = AD   ...(Common)

∠BAD = ∠CAD   ...(Since AB = AC, angles opposite to these sides are equal)

8. By SAS (Side-Angle-Side) congruence criterion, triangles BAD and CAD are congruent:

ΔBAD ≅ ΔCAD

9. Hence proved.