Q12 of 27 Page 289

In the triangle ABC, A is right angle, if CD is a median, let us prove that, BC2 = CD2 + 3AD2

Given: ∠A is right angle.


CD is median.



To prove: BC2 = CD2 + 3AD2


Proof:


We know that median divides the side into two halves.


So,


AB = 2AD = 2BD ………… (1)


By applying Pythagoras theorem in ΔADC, we have,


CD2 = AD2 + AC2


⇒ AC2 = CD2 - AD2 ……… (2)


By applying Pythagoras theorem in ΔABC, we have,


BC2 = AB2 + AC2 ………… (3)


Substituting from eqn. (1) and (2) into (3) gives,


BC2 = (2AD)2 + (CD2 - AD2)


⇒ BC2 = 4AD2 + CD2 - AD2


⇒ BC2 = CD2 + 3AD2


Hence proved.


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