Question

(1)
In Figure, $AD$ is a median of a triangle $ABC$ and $AM\perp BC.$
Prove that :
$\left(i\right)A{C}^2=A{D}^2+BC\times DM+{\left(\frac{BC}{2}\right)}^2$



(2)
In Figure, $AD$ is a median of a triangle $ABC$ and $AM\perp BC.$
Prove that : $\left(ii\right)A{B}^2=A{D}^2-BC\times DM+{\left(\frac{BC}{2}\right)}^2$
(3)
In Figure, $AD$ is a median of a triangle $ABC$ and $AM\perp BC.$
Prove that : $\left(iii\right)A{C}^2+A{B}^2=2A{D}^2+\frac{1}{2}B{C}^2$

Answer 1

(1)

Using Pythagoras Therom in $\Delta AMD\&\Delta AMC$
In $\Delta AMD$
On applying Pythagoras theorem,
$A{D}^2=A{M}^2+M{D}^2.......\left(1\right)$

In $\Delta AMC$
On applying Pythagoras theorem,
$A{C}^2=A{M}^2+M{C}^2$
$A{C}^2=A{M}^2+(MD+DC{)}^2$
$A{C}^2=A{M}^2+M{D}^2+D{C}^2+2\left(MD\right)\left(DC\right)$

Replacing $MC$ in $\Delta AMC$
In $\Delta AMC$
$A{C}^2=A{M}^2+M{C}^2$
$A{C}^2=A{M}^2+(MD+DC{)}^2$
$A{C}^2=A{M}^2+M{D}^2+D{C}^2+2\left(MD\right)\left(DC\right)$
From (1)
$A{C}^2=A{D}^2+D{C}^2+2\left(MD\right)\left(DC\right)......\left(2\right)$

As $AD$ is median, $D$ is the midpoint of $BC$
Substitude $DC=\frac{BC}{2}\text{in}\left(2\right)$
$\therefore A{C}^2=A{D}^2+{\left(\frac{BC}{2}\right)}^2+\left(MD\right)\left(BC\right)$
Hence, proved.

(2)

Using Pythagoras Theorem in $\Delta AMB$ and $\Delta AMD$
In $\Delta AMB$
On applying Pythagoras theorem,
$A{B}^2=A{M}^2+B{M}^2.....\left(1\right)$

In $\Delta AMD$
On applying Pythagoras theorem,
$A{D}^2=A{M}^2+M{D}^2......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, equation $\left(1\right)$ can be written as,
$A{B}^2=(A{D}^2-D{M}^2)+B{M}^2$
$A{B}^2=(A{D}^2-D{M}^2)+(BD-DM{)}^2$ (\because BM=BD-DM)
$A{B}^2=A{D}^2-D{M}^2+B{D}^2+D{M}^2-2\left(BD\right)\left(DM\right)$
$A{B}^2=A{D}^2+B{D}^2-2\left(BD\right)\left(DM\right).....\left(1\right)$

On substituting $BD$
Substitute $BD=\frac{BC}{2}\text{in}\left(3\right)$
$\therefore A{B}^2=A{D}^2-BC\times DM+{\left(\frac{BC}{2}\right)}^2$
Hence Proved.

(3)

Applying Pythagoras Theorem in $\Delta AMB$ and $\Delta AMC$
In $\Delta AMB$
On applying Pythagoras theorem,
$A{B}^2=A{M}^2+B{M}^2.....\left(1\right)$

In $\Delta AMC$
On applying Pythagoras theorem,
$A{C}^2=A{M}^2+M{C}^2........\left(2\right)$

Adding equation (1) &(2).
$A{C}^2+A{B}^2=2A{M}^2+B{M}^2+M{C}^2$
Substitute BM=BD-DM and MC=MD+DC
$A{C}^2+A{B}^2=2A{M}^2+(BD-DM{)}^2+(MD+DC{)}^2$
$A{C}^2+A{B}^2=2A{M}^2+B{D}^2+D{M}^2-2\left(BD\right)\left(DM\right)+M{D}^2+D{C}^2+2\left(MD\right)\left(DC\right)$

Substituting $BD$ and $DC$ Substitude$BD=DC=\frac{BC}{2}$ in the above equation,

$A{C}^2+A{B}^2=2A{M}^2+{\left(\frac{BC}{2}\right)}^2+2D{M}^2-2\left(\frac{BC}{2}\right)\left(DM\right)+M{D}^2+{\left(\frac{BC}{2}\right)}^2+2\left(MD\right)\left(\frac{BC}{2}\right)$

$A{C}^2+A{B}^2=A{D}^2+{\left(\frac{BC}{2}\right)}^2+A{M}^2-\left(BC\right)\left(DM\right)+(MD+DC{)}^2$

$A{C}^2+A{B}^2=A{D}^2+{\left(\frac{BC}{2}\right)}^2+A{M}^2-\left(BC\right)\left(DM\right)+(CM{)}^2$

$A{C}^2+A{B}^2=A{D}^2+{\left(\frac{BC}{2}\right)}^2+A{C}^2-\left(BC\right)\left(DM\right)$
$\therefore A{B}^2=A{D}^2+{\left(\frac{BC}{2}\right)}^2-\left(BC\right)\left(DM\right)$

Hence Proved.