In the figure (1) given below, AD = BD = DC and ∠ACD = 35°. Show that

(i) AC > DC

(ii) AB > AD.

In the figure (2) given below, prove that

(i) x + y = 90°

(ii) z = 90°

(iii) AB = BC.

In the adjoining figure, ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. 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 the adjoining figure, AP ⊥ l and PR > PQ. Show that AR > AQ.