Q3 of 49 Page 61

Let ABC be a triangle such that ∠B = 70° and ∠C = 40°. Suppose D is a point on BC such that AB = AD. Prove that AB > CD.

Capture.JPG


In Δ ABC


AB = AD


∠B = 700


∠ADB = 700(Base angles of an isosceles triangle)


∠ADC = 1800-700 = 1100(Angles on a straight line)


∠C = 400


∠DAC = 1800-(1100 + 400) = 300


Comparing ∠ACD and ∠DAC we find


∠ACD>∠DAC


Since the side opposite to the largest angle is the largest so


AB > CD


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