Question

∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that

(i) ∆ABD ≅ ∆ACD 

(ii) ∆ABP≅ ∆ACP 

(iii) AP bisects ∠A as well as ∠D. 

(iv) AP is the perpendicular bisector of BC.

Answer 1

Given Problem: 

In the triangles \( \triangle ABC \) and \( \triangle DBC \), we are given that both are isosceles with the same base \( BC \), and the vertices \( A \) and \( D \) are on the same side of \( BC \). The point \( P \) is the intersection of the extended line \( AD \) with \( BC \). We need to prove the following:

1. \( \triangle ABD \cong \triangle ACD \)
2. \( \triangle ABP \cong \triangle ACP \)
3. \( AP \) bisects \( \angle A \) as well as \( \angle D \)
4. \( AP \) is the perpendicular bisector of \( BC \)

Proof:

(i) Proof that \( \triangle ABD \cong \triangle ACD \):

Since both \( \triangle ABC \) and \( \triangle DBC \) are isosceles with \( AB = AC \) and \( DB = DC \), we have:

• \( AB = AC \) (Given)
• \( BD = CD \) (Given)
• \( AD = AD \) (Common side)

By the SAS (Side-Angle-Side) congruence criterion, we can conclude that:

\( \triangle ABD \cong \triangle ACD \)

(ii) Proof that \( \triangle ABP \cong \triangle ACP \):

From the congruence of \( \triangle ABD \) and \( \triangle ACD \), we know that:

• \( AB = AC \)
• \( AD = AD \)
• \( \angle ABD = \angle ACD \) (corresponding angles in congruent triangles)

Now, since \( AD \) is extended to \( P \), and \( P \) lies on the base \( BC \), it follows that:

• \( \triangle ABP \cong \triangle ACP \) by SSS (Side-Side-Side) congruence.

(iii) Proof that \( AP \) bisects \( \angle A \) and \( \angle D \):

From the congruence of \( \triangle ABP \) and \( \triangle ACP \), we have:

• \( \angle ABP = \angle ACP \)
• Since \( AP \) is common to both triangles, it divides \( \angle A \) and \( \angle D \) into two equal parts, proving that \( AP \) bisects both angles.

Hence, \( AP \) bisects \( \angle A \) and \( \angle D \).

(iv) Proof that \( AP \) is the perpendicular bisector of \( BC \):

Since \( \triangle ABP \cong \triangle ACP \), the corresponding angles \( \angle ABP \) and \( \angle ACP \) are equal. Furthermore, the line \( AP \) divides the base \( BC \) into two equal segments, proving that \( AP \) is the perpendicular bisector of \( BC \).

Therefore, \( AP \) is the perpendicular bisector of \( BC \).

Conclusion:

By proving these four parts, we have shown that:

• \( \triangle ABD \cong \triangle ACD \)
• \( \triangle ABP \cong \triangle ACP \)
• \( AP \) bisects both \( \angle A \) and \( \angle D \)
• \( AP \) is the perpendicular bisector of \( BC \)