In triangle $A B C, A D$ is perpendicular to $B C$ .Prove that:
$A B^{2} - A C^{2} = B D^{2} - C D^{2}$

Open in App

Solution

Since triangles $A B D$ and $A C D$ are right angled at $D$ .
Therefore, $A B^{2} = A D^{2} + B D^{2}$ ......................1
And, $A C^{2} = A D^{2} + C D^{2}$ ........................2
Subtracting 2 from 1 we get
$A B^{2} - A C^{2} = B D^{2} - C D^{2}$

Suggest Corrections

Similar questions

△ A B C

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

A D ⊥ B C

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

If $A D ⊥ B C$ , prove that ${A B}^{2} + {C D}^{2} = {B D}^{2} + {A C}^{2}$ . [2 MARKS]

A D ⊥ B C

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

Join BYJU'S Learning Program