Question

in a Δ abc ac ad d is the mid point of bc and ae perpendicular BC prove that AC^2 =AD^2 BC *DE 1/4BC^2

Answer 1

Hey ! Folk here's your answer

AEB is a right angled triangle.
Therefore, by Pythagoras theorem,
AB 2= AE2 + BE2 --------------- (1)
Now, AED is a right angled triangle.
Therefore,
=> AE2 + ED2= AD2 (Pythagoras theorem)
=> AE2 = AD2 - ED2 -------------- (2)
Now, BE = BD - ED ---------------- (3)
Substituting (2) and (3) in (1), we get,
AB2= AD2 - ED2 + (BD - ED)2
=AD2 – ED2 + BD2 -2BD*ED + ED2
= AD2 – ED2 + BD2 -2BD*ED + ED2
Now, BD = BC/2 (since, D is midpoint)
Therefore,
AB2= AD2 + BD2 -2BD*ED
=AD2 + (BC/2)2 – 2(BC/2)*ED
= AD2 + (BC/2)2 – 2(BC/2)*ED
= AD2 – BC*DE + ¼ BC2
HENCE PROVED
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