In triangle ABC, angle A=60. Prove that BC2=AB2+AC2- AB.AC
Given: ΔABC, ∠A = 60°
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To Prove: BC2 = AB2 + AC2 – AB × AC
Construction: Draw CD ⊥ AB
Proof: In right ΔBCD,
BC2 = CD2 + BD2 … (1) (Pythagoras Theorem)
In right ΔACD,
AC2 = AD2 + CD2 (Pythagoras Theorem)
⇒ AC2 – AD2 = CD2 … (2)
From (1) and (2),
BC2 = AC2 – AD2 + BD2
⇒BC2 = AC2 – AD2 + (AB – AD)2
⇒ BC2 = AC2 – AD2 + AB2 + AD2 – 2AB × AD
⇒ BC2 = AC2 + AB2 – 2AB × AD … (3)
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