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Maths-

General

Easy

Question

$text In an isosceles Triangle end text A B C comma A B equals A C text and end text D text is a point on end text B C text produced. Prove that end text A D squared equals A C squared plus B D. C D$

hint Hint:

Hint :- Using pythagoras theorem in triangle AEC and AED we find AC2 and AD2 and do $A D squared minus A C squared$ equation to get the required result.

The correct answer is: Hence proved

Aim :- Prove that $A D squared equals A C squared plus B D times C D$
Explanation(proof ) :-
In triangles ABE and ACE ,
∠AEB = ∠AED = 90⁰ (Right angle)
AB = AC (given) (Hypotenuse)
AE = AE (common side)
By RHS , ΔAEB ≅ ΔAEC
By congruence , BE = EC
Applying pythagoras theorem in triangle AEC we get,
$A C squared equals E C squared plus A E squared$ — Eq1
Applying pythagoras theorem in triangle AED we get,
$A D squared equals E D squared plus A E squared$ — Eq2

Doing Eq2-Eq1 we get , $A D squared minus A C squared equals E D squared plus A E squared minus A E squared minus E C squared$
$not stretchy rightwards double arrow A D squared minus A C squared equals E D squared minus E C squared equals left parenthesis E D plus E C right parenthesis left parenthesis E D minus E C right parenthesis$
Substitute ED-EC = CD
$open not stretchy rightwards double arrow A D squared minus A C squared equals left parenthesis C D right parenthesis left parenthesis E D plus E C right parenthesis text [substitute end text EC equals BE text from above result end text close square brackets$
$not stretchy rightwards double arrow A D squared minus A C squared equals left parenthesis C D right parenthesis left parenthesis E D plus B E right parenthesis$
$A D squared minus A C squared equals B D times C D$
$therefore A D squared equals A C squared plus B D times C D$
Hence proved

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