ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig. 7.17). Prove that
(i) ∆ ABD ≅ ∆ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC.

We are given two triangles \( \triangle ABD \) and \( \triangle BAC \). Our objective is to prove that these two triangles are congruent and to demonstrate the equality of certain sides and angles.
Step 1: Understanding the given information:
In the triangle \( \triangle ABD \), we are given the following facts:
Step 2: Applying the SAS (Side-Angle-Side) Congruence Rule:
We now use the SAS congruence rule to prove that the two triangles are congruent. The SAS rule states that if two sides and the included angle of one triangle are respectively equal to the corresponding two sides and the included angle of another triangle, then the two triangles are congruent.
In our case:
Therefore, by the SAS congruence rule, we can conclude that:
\[ \triangle ABD \cong \triangle BAC \quad \text{(By SAS congruence rule)} \]
Step 3: Applying CPCT (Corresponding Parts of Congruent Triangles):
Since we have established that the two triangles are congruent, we can apply the principle of CPCT, which states that corresponding parts of congruent triangles are equal. Therefore, from \( \triangle ABD \cong \triangle BAC \), we deduce the following:
Conclusion: By applying the SAS congruence rule and the CPCT property, we have proven that:
This demonstrates that \( \triangle ABD \) and \( \triangle BAC \) are congruent, and certain corresponding parts of these triangles are indeed equal.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A.
∆ 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.

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that:
(i) ∆ AMC ≅ ∆ BMD
(ii) ∠ DBC is a right angle.
(iii) ∆ DBC ≅ ∆ ACB (iv) CM = \(\frac{1}{2}\) AB

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB (see Fig). Show that
(i) ∆ DAP ≅ ∆ EBP
(ii) AD = BE
