Question

In given figure, ABC is a right triangle right angled at B and D is the foot of the perpendicular drawn from B on AC. If $DM\perp BC$ and $DN\perp AB$, prove that

(i) ${DM}^2=DN\times MC$

(ii) ${DN}^2=DM\times AN$

Answer 1

We have,

$AB\perp BC$ and $DM\perp BC$

$\Rightarrow$ $AB||DM$

Similarly, we have

$CB\perp AB$ and $DN\perp AB$

$\Rightarrow$ $CB||DN$

Hence, quadrilateral BMDN is a rectangle.

$\therefore$ $BM=ND$

(i) In $△BMD$, we have

$\angle 1+\angle BMD+\angle 2={180}^0$

$\Rightarrow$ $\angle 1+{90}^0+\angle 2={180}^0$

$\Rightarrow$ $\angle 1+\angle 2={90}^0$

Similarly, in $△DMC$, we have

$\angle 3+\angle 4={90}^0$

Since $BD\perp AC$. Therefore,

$\angle 2+\angle 3={90}^0$

Now, $\angle 1+\angle 2={90}^0$ and $\angle 2+\angle 3={90}^0$

$\Rightarrow$ $\angle 1+\angle 2=\angle 2+\angle 3$

$\Rightarrow$ $\angle 1=\angle 3$

Also, $\angle 3+\angle 4={90}^0$ and $\angle 2+\angle 3={90}^0$

$\Rightarrow$ $\angle 3+\angle 4=\angle 2+\angle 3\Rightarrow \angle 2=\angle 4$

Thus, in ${△}^{\prime }sBMD$ and $DMC$, we have

$\angle 1=\angle 3$ and $\angle 2=\angle 4$

So, by AA-criterion of similarity, we have

$△BMD\sim △DMC$

$\Rightarrow$ $\frac{BM}{DM}=\frac{MD}{MC}$

$\Rightarrow$ $\frac{DN}{DM}=\frac{DM}{MC}$ [$\because BM=ND$]

$\Rightarrow$ ${DM}^2=DN\times MC$ $\left[Henceproved\right]$

(ii) Proceeding as in (i), we can prove that

$△BND\sim △DNA$

$\Rightarrow$ $\frac{BN}{DN}=\frac{ND}{NA}$

$\Rightarrow$ $\frac{DM}{DN}=\frac{DN}{AN}$

$\Rightarrow$ ${DN}^2=DM\times AN$ $\left[Henceproved\right]$