l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ∆ ABC ≅ ∆ CDA.

In \( \triangle ABC \) and \( \triangle CDA \), we have the following:
First, we know that:
\[ \angle BAC = \angle DCA \quad \text{(Alternate interior angles, as } p \parallel q) \]
Also, \( AC = CA \) (Common side).
Next, we observe that:
\[ \angle BCA = \angle DAC \quad \text{(Alternate interior angles, as } l \parallel m) \]
Hence, by the **ASA congruence rule**, we conclude that:
\[ \triangle ABC \cong \triangle CDA \]
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
