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If Two Sides of a Triangle Are Unequal, Then the Angle Opposite to the Greater Side Is Greater Than Angle Opposite to the Smaller Side

D and E are p...

Question

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Then:

A E^{2} - B D^{2} = A B^{2} - D E^{2}

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A E^{2} + B D^{2} = A B^{2} + D E^{2}

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A E^{2} - D E^{2} = A B^{2} - B D^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

A E^{2} + D E^{2} = A B^{2} + B D^{2}

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Solution

The correct options are
B $A E^{2} + B D^{2} = A B^{2} + D E^{2}$
C $A E^{2} - D E^{2} = A B^{2} - B D^{2}$

In $Δ A E C$ ,
$A E^{2} = C E^{2} + A C^{2} - - (1)$

In $Δ B D C$ ,
$B D^{2} = C D^{2} + B C^{2} - - (2)$

Adding (1) and (2), we get
$A E^{2} + B D^{2}$
$= C E^{2} + C D^{2} + A C^{2} + B C^{2}$
But, in $Δ C D E$ ,
$D E^{2} = C D^{2} + C E^{2}$
And, in $Δ A B C$ ,
$A B^{2} = A C^{2} + B C^{2}$

$∴ A E^{2} + B D^{2} = A B^{2} + D E^{2}$
and, $A E^{2} - D E^{2} = A B^{2} - B D^{2}$ (By rearranging terms)

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Similar questions

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE²+ BD² = AB² + DE²

A E^{2} + B D^{2} = A B^{2} + D E^{2}

D

E

C A

C B

△ A B C

C

A E^{2} + B D^{2} = A B^{2} + D E^{2}

{A Q}^{2}

{B P}^{2} = ?

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