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Basic Proportionality Theorem

In triangle ...

Question

In triangle $A B C, A B > A C . E$ is the mid-point of $B C$ and $A D$ is perpendicular to $B C$ . Prove that:
$A B^{2} + A C^{2} = 2 A E^{2} + 2 B E^{2}$

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Solution

In $Δ A B D$ ,
${A B}^{2} = {B D}^{2} + {A D}^{2}$ --(1)

In $Δ A D C$ ,
${A C}^{2} = {C D}^{2} + {A D}^{2}$ --(2)

Adding (1) and (2),
${A B}^{2} + {A C}^{2} = 2 {A D}^{2} + {B D}^{2} + {C D}^{2} . . . . . . . . . . . (8)$

${B D}^{2} = {B E}^{2} + {E D}^{2} + 2 B E . E D$ --(3)
$C D = E C - E D {C D}^{2} = {E C}^{2} + {E D}^{2} - 2 E C . E D$ --(4)

Adding (3) and (4),
${B D}^{2} + {C D}^{2} = {B E}^{2} + {E C}^{2} + 2 {E D}^{2} + 2 B E . E D - 2 E C . E D$
$= {B E}^{2} + {E C}^{2} + 2 {E D}^{2} + 2 E D (B E - E C)$
$= {2 B E}^{2} + 2 {E D}^{2}$ ....[ $∵ B E = E C$ ]

In $Δ A E D$ ,
$A E^{2} = {E D}^{2} + {A D}^{2} {E D}^{2} = A E^{2} - {A D}^{2} {B D}^{2} + {C D}^{2} = {2 B E}^{2} + 2 A E^{2} - {2 A D}^{2} . . . . . . . . . . . . . . . . . . . . . . . . . (5)$

From (5) +(8),
${A B}^{2} + {A C}^{2} = 2 {A D}^{2} + {2 B E}^{2} + 2 A E^{2} - {2 A D}^{2}$
$= {2 B E}^{2} + 2 A E^{2}$

Hence proved.

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