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Pythagoras Theorem

Δ ABC is a ri...

Question

$Δ A B C$ is a right angled triangle in which $∠ B = 90^{\circ} .$ D and E are any point on AB and BC respectively. Prove that $A E^{2} + C D^{2} = A C^{2} + D E^{2} .$

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Solution

In right $Δ A B C, ∠ B = 90^{\circ}$ and D, E are point of AB, BC respectively.
To prove:
$A C^{2} + D E^{2} = A E^{2} + C D^{2}$
In $r i g h t Δ A B C$ by using Pythagoras theorem,
$A C^{2} = A B^{2} + B C^{2}$ ...(i)
In $r i g h t Δ A B E$ by using Pythagoras theorem
$A E^{2} = A B^{2} + B E^{2}$ ...(ii)
In $r i g h t Δ B C D$ by using Pythagoras theorem,
$C D^{2} = B D^{2} + B C^{2}$ ...(iii)
In $r i g h t Δ D B E$ by using Pythagoras theorem,
$D E^{2} = D B^{2} + B E^{2}$ ...(iv)
Adding eq. (i) and eq. (iv)
$A C^{2} + D E^{2} = A B^{2} + B C^{2} + B D^{2} + B E^{2}$
$= A B^{2} + B E^{2} + B C^{2} + B D^{2}$
$A C^{2} + D E^{2} = A E^{2} + C D^{2}$
[From eq. (ii) and eq. (iii)]
Hence Proved.

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