
Let us consider that ABC is an equilateral triangle.
Therefore, AB = BC = AC AB = AC
∠C =∠ B (Angles opposite to equal sides of a triangle are equal)
Also,
Ac = BC
∠B = ∠A (Angles opposite to equal sides of a triangle are equal)
Therefore, we obtain A
= ∠B = ∠C
In ∆ABC,
∠A + ∠B + C = 180°
∠A + ∠A +∠A = 180°
∠3A = 180°
∠A = 60°
∠A = ∠B = ∠C = 60°
Hence, in an equilateral triangle, all interior angles are of measure 60º.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 9.27). Prove that ∠ACP = ∠ QCD

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.14). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∠∆BAD
(iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

(i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)