Question

In Fig., $D$ is a point on hypotenuse $AC$ of $△ABC$, such that $BD\perp AC,DM\perp BC$ and $DN\perp AB.$ Prove that :
(i) $D{M}^2=DN.MC$
(ii) $D{N}^2=DM.AN$

Answer 1

i) In $\Delta ABC$,

$DN\perp AB$ and $BC\perp AB$

So, $DN\parallel BC$ ...(1)

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

So, $DM\parallel AB$ ...(2)

From (1) and (2),

$\square DMBN$ is a rectangle.

$\therefore BM=DN$

In $\Delta BMD$,

$\angle M+\angle BDM+\angle DBM={180}^o$

$\Rightarrow \angle BDM+\angle DBM={90}^o$ ...(1)

Similarly, in $\Delta DMC$,

$\angle CDM+\angle MCD={90}^o$ ...(2)

We know, $BD\perp AC$ ....given

$\therefore \angle BDM+\angle MDC={90}^o$ ..(3)

From (1) and (3), we get

$\angle BDM+\angle DBM=$$\angle BDM+\angle MDC$

$\therefore \angle DBM=\angle MDC$ ...(4)

Similarly, $\angle BDM=\angle MCD$ ...(5)

In $\Delta BMD$ and $\Delta DMC$,

$\angle BMD=\angle DMC$ ...Each ${90}^o$

$\angle DBM=\angle MDC$ ...From (4)

$\angle BDM=\angle MCD$ ...From (5)

$\Delta BMD\sim$$\Delta DMC$ ....AAA test of similarity

$\therefore \frac{BM}{DM}=\frac{MD}{MC}$ ....C.S.S.T.

$\therefore \frac{DN}{DM}=\frac{DM}{MC}$ ...$\because BM=ND$

$\Rightarrow D{M}^2=DN\times MC$

ii) Similarly, we can prove $\Delta DNB\sim \Delta DNA$

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

$\frac{DM}{DN}=\frac{DN}{AN}$ ...$[\because BN=DM]$

$D{N}^2=DM\times AN$