in ∆abc ,d is a point on side ac such that ab÷bc =ad÷dc then prove that angle abd ~= to angle cbd
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Consider a ∆ABC in which ∠B is a right angle and ∠C is double of ∠A
To Prove: AC = 2BC
Construction: produce CB upto D such that BC = BD & join AD
Proof:
In ∆ ABD & ∆ ABC
BD= BC. (By construction)
∠ABD = ∠ABC ( 90°)
AB = AB (Common)
∆ABD ∆ABC. ( BY SAS)
AD= AC. (CPCT)........(1)
∠DAB = ∠CAB = P ( Let) (By CPCT).......(2)
∠DAC = ∠DAB + ∠CAB
∠DAC= P + P = 2P. ( From eq 2)
∠DAC = ∠ACB ( ∠ACB= 2P)
DC = AD
(Since, sides opposite to equal angles of a triangle are equal)
2BC = AD= AC ( BC= DB from eq 1)
Hence, AC= 2BC
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