Question

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Answer 1

In right DADB, AB2= AD2+ BD2[By Pythagoras theorem] – (1) In right DADC, AC2= AD2+ CD2[By Pythagoras theorem] – (2) Subtract (1) from (2), we get AC2– AB2= AD2+ CD2– AD2– BD2 AC2– AB2= CD2– BD2�--- (3)Given BD =1/3 CD From the figure, BD + DC = BC ⇒ (1/3)CD + CD = BC ⇒ (4/3)CD = BC ∴ CD = 3BC/4 ⇒ BD = BC/4 Equation (3) becomes, AC2– AB2= (3BC/4)2– (BC/4)2 �=(8/16)BC2 ⇒ AC2– AB2= (1/2) BC2 ⇒ 2AC2– 2AB2= BC2 ⇒ 2AC2= 2AB2+ BC2