Question

in a triangle abc ac ab d is the midpoint of bc and AD perpendicular bisector of BC. prove that (1) AC²=AD²+BC. DE +¼BC² (2) AB²=AD² - BC. DC+¼BC² . Let tell me the ans ​

Answer 1

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Answer 2

AEB is a right angled triangle.

Therefore, by Pythagoras theorem,

AB 2= AE2 + BE2

(1) (3) AB2= AD2 - ED2 + (BD - ED)2 = AD2 - ED2 + BD2 -2BD*ED + ED2

Now, AED is a right angled triangle.

Therefore,

=> AE2 + ED2= AD2 (Pythagoras theorem) (2)

=> AE2 = AD2 - ED2

Now, BE = BRED

Substituting (2) and (3) in (1), we get,

=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 + 4 BC2