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

Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

In an isoscel...

Question

In an isosceles triangle $A B C, A B = A C$ and $D$ is a point on $B C$ produced. $A M$ is perpendicular on $B C$ .Prove that $A D^{2} = A C^{2} + B D . C D$

Open in App

Solution

In $△ A M D$ ,
${A D}^{2} = {A M}^{2} + {M D}^{2}$ --(1)
In $△ A M C,$
${A C}^{2} = {A M}^{2} + {M C}^{2}$ --(2)
$B D = B C + C D = 2 M C + C D$
$M D = M C + C D$
${M D}^{2} = {M C}^{2} + {C D}^{2} + 2 M C . C D$ --(3)
From (1), ${A M}^{2} = {A D}^{2} - {M D}^{2}$

Adding eq.(1) and eq.(2), we get
${A D}^{2} + {A C}^{2} = 2 {A M}^{2} + {M C}^{2} + {M D}^{2} {A D}^{2} + {A C}^{2} = 2 {A M}^{2} + {M C}^{2} + {M C}^{2} + {C D}^{2} + 2 M C . C D$
$= 2 {A M}^{2} + {2 M C}^{2} + {C D}^{2} + 2 M C . C D$

Substituting for ${A M}^{2} = {A D}^{2} - {M D}^{2}$ in the above step,

${A D}^{2} + {A C}^{2} = {2 M C}^{2} + {C D}^{2} + 2 M C . C D + 2 ({A D}^{2} - {M D}^{2})$
$= {2 M C}^{2} + {C D}^{2} + 2 M C . C D + 2 ({A D}^{2} - {M C}^{2} - {C D}^{2} - 2 M C . C D)$
$= 2 {A D}^{2} + {2 M C}^{2} + {C D}^{2} + 2 M C . C D - 2 {M C}^{2} - 2 {C D}^{2} - 4 M C . C D$
$∴ {A D}^{2} + {A C}^{2} = 2 {A D}^{2} - {C D}^{2} - 2 M C . C D$

$B D . D C = (2 M C + C D) . D C B D . D C = 2 M C . C D + {C D}^{2}$

Substituting in ${A D}^{2} + {A C}^{2} = 2 {A D}^{2} - {C D}^{2} - 2 M C . C D$ , we get

${A D}^{2} + {A C}^{2} = 2 {A D}^{2} - {C D}^{2} - 2 M C . C D$
${A D}^{2} + {A C}^{2} = 2 {A D}^{2} - B D . D C$
${A C}^{2} = {A D}^{2} - B D . D C$
${A D}^{2} = {A C}^{2} + B D . D C$
Hence proved.

Suggest Corrections

Similar questions

A B C, A B = A C

D

B C

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

A B C

A B = B C

D

B C

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

Δ

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

△ A B C

A B = A C

D

B C

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

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

Join BYJU'S Learning Program