MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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 $A E^{2} + C D^{2} = A C^{2} + D E^{2} .$

Open in App

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.

Suggest Corrections

thumbs-up

1

Similar questions

Q. In given figure ABC is a right-angled triangle at B. Let D and E be any points on AB and BC respectively. Prove that

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

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

Q. Question 13
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that

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

D

E

are points on the sides

C A

C B

repesctively of a

△ A B C

, right angled at

C

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

Q. ΔABC is right-angled at B and D is the mid-point of BC.
Prove that:

A C^{2} = (4 A D^{2} - 3 A B^{2}) .

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Pythogoras Theorem

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Pythagoras Theorem

Standard X Mathematics

Join BYJU'S Learning Program

CrossIcon