Question

In given Figure $D$ is a point on hypotenuse $AC$ of $△ABC$, such that $BD\perp AC$, $DM\perp BC$ and $DN\perp AB$.

Prove that: $D{M}^2=DN.MC$

Answer 1

To prove $D{M}^2=DN.MC$

Given that: $D$ is a point on hypotenuse $AC$ of $△ABC$, such that $BD\perp AC$, $DM\perp BC$ and $DN\perp AB$

Proof:

Step 1: Verify that $△DNA~△BND$

Let us join Point $D$ and $B$ by a straight line.

As we know $DN\parallel CB$, $DM\parallel AB$ and $\angle B=90°$.

$DMBN$ is a rectangle.

Therefore, $DN=MB$ and $DM=NB$

Also, $\angle CBD=90°$

From figure,

$\angle 2+\angle 3=90°$--------- $\left(A\right)$

From $△CDM$,

$\angle 1+\angle 2+\angle DMC=180°$

$\angle 1+\angle 2+90°=180°$

$\angle 1+\angle 2=180°-90°$

$\angle 1+\angle 2=90°$ -------- $\left(B\right)$

From $△DMB$,

$\angle 3+\angle 4+\angle DMB=180°$

$\angle 3+\angle 4+90°=180°$

$\angle 3+\angle 4=180°-90°$

$\angle 3+\angle 4=90°$ ---------- $\left(C\right)$

Equating equation $\left(A\right)$ and $\left(B\right)$

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

$\angle 1=\angle 3$

Equating equation $\left(A\right)$ and $\left(C\right)$

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

$\angle 2=\angle 4$

Therefore, By $AA$ similarity criterion $△BMD~△DMC$

Step 2: Use the property of similarity of two triangles

$$\begin{gathered} \frac{BM}{DM}=\frac{DM}{MC} \\ D{M}^2=BM.MC \end{gathered}$$

Since, $BM=DN$

$\mathbf{\therefore }\mathit{D}{\mathit{M}}^{\mathbf{2}}\mathbf{=}\mathit{D}\mathit{N}\mathbf{}\mathbf{.}\mathbf{}\mathit{M}\mathit{C}$

Hence proved.