Question

In triangle ABC, AD is median and AE is perpendicular to BC, prove that -
AB^2 + AC^2 = 2AD^2 + 1/2BC^2

Answer 1

Using Pythagoras therorom
AD^2+BD^2=AB^2......... (1)
AD^2+DC^2=AC^2.........(2)
now adding equations 1 and 2 we get....
AB^2+AC^2=2AD^2+1/2BC^2

Answer 2

Step 1:  Kindly refer diagram.

Step 2: BD=DC (since AD is median)

In angle AED

$\mathrm{AD}^{2}=\mathrm{A} \mathrm{E}^{2}+\mathrm{D} \mathrm{E}^{2}$[PYTHAGORAS Theorem]

$\mathrm{AE}^{2}=\mathrm{AD}^{2}-\mathrm{D} \mathrm{E}^{2}$……….(1)

Step 3:

IN ANGLE AEB,

$\mathrm{AB}^{2}=\mathrm{AE}^{2}+\mathrm{BE}^{2}$

= $\mathrm{AD}^{2}-\mathrm{D} \mathrm{E}^{2}+\mathrm{BE}^{2}$[Substitute value from from equation (1)]

=$(\mathrm{BD}+\mathrm{DE})^{2+} \mathrm{AD}^{2}-\mathrm{D} \mathrm{E}^{2}$ [BE=BD+DE]

$\begin{array}{l}{=\mathrm{BD}^{2}+\mathrm{DE}^{2}+2 \mathrm{BD}^{*} \mathrm{DE}+\mathrm{AD}^{2}-\mathrm{DE}^{2}} \\ {=\mathrm{BD}^{2}+\mathrm{AD}^{2}+2 * \mathrm{BD}^{*} \mathrm{DE} \ldots \ldots . .(2)}\end{array}$

Step 4:

IN ANGLE AED,

$\mathrm{AD}^{2}=\mathrm{A} \mathrm{E}^{2}+\mathrm{D} \mathrm{E}^{2}$ [Pythagoras Theorem]

$\mathrm{AE}^{2}=\mathrm{AD}^{2}-\mathrm{DE}^{2}$  ……..(3)

Step 5:

IN ANGLE AEC,

$\mathrm{AC}^{2}=\mathrm{AE}^{2}+\mathrm{EC}^{2}$

= $\mathrm{AD}^{2}-\mathrm{DE}^{2}+\mathrm{EC}^{2}$ [Substitute value from equation 3]

 

Step 6:

GIVEN BD=DC

THEREFORE $\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}-2 * \mathrm{BD} * \mathrm{DE}$……(4)

ADDING EQUATION (2) AND (4) WE GET

Step 7:

$\begin{array}{l}{\mathrm{AB}^{2}+\mathrm{AC}^{2}=\mathrm{BD}^{2}+\mathrm{AD}^{2}+2 \mathrm{BD}^{*} \mathrm{DE}+\mathrm{AD}^{2}+\mathrm{BD}^{2}-2^{*} \mathrm{BD}^{*} \mathrm{DE}} \\ {\mathrm{AB}^{2}+\mathrm{AC}^{2}=2\left(\mathrm{AD}^{2}+\mathrm{BD}^{2}\right)} \\ {\mathrm{AB}^{2}+\mathrm{AC}^{2}=2\left(\mathrm{AD}^{2}+(\mathrm{BC} / 2)^{2}\right)}\end{array}$

HENCE PROVED