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Question

Question: If D is the mid-point of side BC of a triangle ABC and AD is perpendicular to AC, then...

If D is the mid-point of side BC of a triangle ABC and AD is perpendicular to AC, then

A

3b2 = a2 – c2

B

3a2 = b2 – 3c2

C

b2 = a2 – c2

D

a2+ b2= 5c2

Answer

3b2 = a2 – c2

Explanation

Solution

From the right-angled triangle CAD, we have

cos C =

Ž 2ba=a2+b2c22ab\frac { 2 b } { a } = \frac { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } } { 2 a b }

Ž a2 + b2 – c2 = 4b2

Ž a2 – c2 = 3b2