Question

Plzz solve this problem.. I am Not getting the answer​

Answer 1

Given :

• In ΔABC, $\angle{B\:=\:90\:^\circ}$
• D is the midpoint of BC.

To prove :

• $\bold{AC^2\:=\:AD^2\:+\:3\:CD^2}$

Proof :

In Δ ABD, $\angle{B\:=\:90\:^\circ}$

•°• By using pythagoras theorem,

$\sf{AD^2\:=\:AB^2\:+\:BD^2\:\:\:(1)}$

Now, consider the ΔABC.

D is the midpoint of BC.

$\sf{\big[Given\big]}$

•°• $\bold{BD\:=\:DC\:\:\:(2)}$

$\bold{\angle{B\:=\:90\:^\circ\:\:\:(Given)}}$

•°• Using pythagoras theorem,

$\sf{AC^2\:=\:AB^2\:+\:BC^2\:}$

$\sf{AC^2\:=\:AB^2\:+\:(BD\:+\:DC)^2}$

$\sf{\big[\because\:BC\:=\:BD\:+\:DC\:\:\:(B-D-C)\big]}$

$\sf{AC^2\:=\:AB^2\:+\:BD^2\:+\:2BD\:\times\:DC\:+\:DC^2}$

$\bold{\big[Using\:the\:identity\:(a+b)^2\:=\:a^2\:+\:2ab\:+\:b^2\big]}$

$\sf{AC^2\:=\:AB^2\:+\:CD^2\:+\:2CD\:\times\:CD\:+\:CD^2}$

$\bold{\big[From\:equation(2)\:BD\:=\:DC\big]}$

$\sf{AC^2\:=\:AB^2\:+\:CD^2\:+\:2CD^2\:+\:CD^2}$

$\sf{AC^2\:=\:AB^2\:+\:CD^2\:+\:CD^2\:+\:2CD^2}$

$\sf{AC^2\:=\:AB^2\:+\:2CD^2\:+\:2CD^2}$

$\sf{AC^2\:=\:AB^2\:+\:4CD^2\:\:\:(3)}$

Now, subtract equation (1) from (3),

$\sf{AB^2+4CD^2\:-(AB^2+CD^2)\:=\:AC^2-AD^2}$

$\bold{\big[Using\:equation\:(2)\big]}$

$\sf{\cancel{AB^2}+4CD^2\cancel{-AB^2}-CD^2=AC^2-AD^2}$

$\sf{4CD^2-CD^2=AC^2-AD^2}$

$\sf{3CD^2\:=\:AC^2-AD^2}$

$\sf{3CD^2+AD^2=AC^2}$

$\large{\boxed{\bold{AC^2\:=\:AD^2\:+\:3CD^2}}}$