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ABC is a triangle. D is the middle point of BC. If AD is perpendicular to AC, then prove that Cos A cos C = 2(c 2 – a 2 ) / 3ac

ABC is a triangle. D is the middle point of BC. If AD is perpendicular to AC, then prove that
Cos A cos C = 2(c2 – a2) / 3ac 

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
9 years ago
Hello Student,
Please find the answer to your question
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In ∆ ACD, cos C = b/a/2 = 2b/a . . . . . . . . . . . . . . . . . . . . . . . . (1)
In ∆ ABC, cos C = a2 + b2 + c2/2ab . . . . . . . . . . . . . . . . . . . . . . . . (2)
From (1) and (2),
2b/a = a2 + b2 - c2/2ab ⇒ b2 = 1/3 (a2 – c2) . . . . . . . . . . . . . . (3)
Also cos A = b2 + c2 – a2/2bc
∴ cos A cos C = b2 + c2 – a2/2bc x 2b/a = b2 + c2 – a2/ac
= \frac{\frac{1}{3}(a^2-c^2)+(c^2-a^2)}{ac}= 2(c2 – a2)/3ac

Thanks
Aditi Chauhan
askIITians Faculty

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