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

Open in App

Solution

Concept: 1 Mark
Application: 1 Mark

In $Δ A D C$ , we have

${A C}^{2} = {A D}^{2} + {C D}^{2} . . . . . (1)$ [Pythagoras Theorem]

In $Δ A D B$ , we have

${A B}^{2} = {A D}^{2} + {B D}^{2} . . . . . . . (2)$ [Pythagoras Theorem]

Subtracting (1) from (2), we have

${A B}^{2} - {A C}^{2} = {B D}^{2} - {C D}^{2}$

or, ${A B}^{2} + {C D}^{2} = {B D}^{2} + {A C}^{2}$

Suggest Corrections

Similar questions

A D ⊥ B C

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

A D ⊥ B C

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

△ A B C

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

A B C, A D

B C

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

A D ⊥

B C, B - D - C

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

Join BYJU'S Learning Program