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Maths-

General

Easy

Question

The perpendicular AD on the base BC of Triangle ABC intersects BC in D such that BD = 3CD. Prove that
$2 A B squared equals 2 A C squared plus B C squared$

hint Hint:

Hint :- using pythagoras theorem Find AC2 and AB2 subtract both the equation and substitute given conditions

The correct answer is: Hence proved

Aim :- Prove that $2 A B squared equals 2 A C squared plus B C squared$
Explanation(proof ) :-
As BD = 3 CD ;We get BC = BD +CD = 4 CD
Applying pythagoras theorem in ΔACD ,We get
$A C squared equals A D squared plus C D squared$ — Eq1
Applying pythagoras theorem in ΔABD ,We get
$A B squared equals B D squared plus A D squared not stretchy rightwards double arrow A D squared equals A B squared minus B D squared$ —Eq2

Substitute Eq2 in Eq1 ,
$A C squared equals A B squared minus B D squared plus C D squared not stretchy rightwards double arrow A C squared equals A B squared plus C D squared minus B D squared$
As BD = 3 CD ;We get $A C squared equals A B squared plus C D squared minus 9 C D squared$
$A C squared equals A B squared minus 8 C D squared not stretchy rightwards double arrow 8 C D squared plus A C squared equals A B squared$
Multiplying with 2 on both sides
$left parenthesis 4 C D right parenthesis squared plus 2 A C squared equals 2 A B squared AsBC equals 4 CD$
$therefore 2 A B squared equals 2 A C squared plus B C squared$
Hence proved

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