In the given fig, AD is a median of a triangle ABC and AM ⊥ B

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235.

In the given fig, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
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(iii)   AC squared plus AB squared space equals space 2 AD squared plus 1 half BC squared.




(i) In right triangle ACM, we have
AC2 = AM2 + MC2 [Using Pythagoras theorem]
equals space AM squared plus left parenthesis MD plus DC right parenthesis squared
equals space AM squared plus MD squared plus DC squared plus 2 MD. DC
equals space AD squared plus open parentheses 1 half BC close parentheses squared plus 2 DM.1 half BC
equals AD squared plus 1 fourth BC squared plus BC. DM space space space space space space space space space space... left parenthesis straight i right parenthesis
(ii) In right triange ABM, we have
AB2 = AM2 + BM2 [Using Pythagoras theorem]
= AM2 + (BD - MD)2
= AM2 + BD2 + MD2 - 2BD.MD
= (AM2 + MD2) + BD2 - 2BD.MD
AD squared plus open parentheses 1 half BC close parentheses squared minus 2 cross times 1 half BC. DM
equals AD squared plus 1 fourth BC squared minus BC. DM space space space space... left parenthesis ii right parenthesis
Adding (i) & (ii), we get
AC squared plus AB squared equals 2 AD squared plus 1 half BC squared.
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