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Maths-

General

Easy

Question

$straight angle A$ of an isosceles triangle ABC is acute in which $A B equals A C text and end text B D perpendicular A C$ and . Prove that
$B C squared equals 2 cross times A C cross times C D$

The correct answer is: not stretchy rightwards double arrow B C squared equals 2 ∗ A C ∗ C D

Solution :-
Aim :- prove $B C squared equals 2 ∗ A C ∗ C D$
Hint :- Using pythagoras theorem, find the length of side AB and BC .
Substitute the RHS of $B C squared equals C D squared plus B D squared$ in terms of AC and CD Explanation(proof
Using pythagoras theorem,in triangle BDC
$B C squared equals C D squared plus B D squared$ — Eq1
Using pythagoras theorem,in triangle BDA $A B squared equals B D squared plus A D squared not stretchy rightwards double arrow B D squared equals A B squared minus A D squared$ — Eq2
Substitute Eq2 in Eq1
$B C squared equals A B squared minus A D squared plus C D squared$
Substitute AD = AC - CD (from diagram)
$B C squared equals A B squared minus left parenthesis A C minus C D right parenthesis squared plus C D squared not stretchy rightwards double arrow B C squared$
$equals A B squared minus A C squared minus C D squared plus 2 ∗ A C ∗ C D plus C D squared$
$text As end text A B equals A C left parenthesis text given end text right parenthesis text we get end text comma B C squared equals A C squared minus A C squared minus C D squared plus 2 ∗ A C ∗ C D plus C D squared$
$not stretchy rightwards double arrow B C squared equals 2 ∗ A C ∗ C D$
Hence proved

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