Question

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

Answer 1

$\left(i\right)$ In $△BDC$

$DM\perp BC$

$△BMD\sim △DMC....\left(1\right)$ [If a perpendicular is drawn the vertex of the right angle of a right triangle to the hypotenuse, the triangle on both sides of the perpendicular are similar to the whole triangle and to each other]

In $△BDA$

$DN\perp AB$

$△AND\sim △DNB....\left(2\right)$ [If a perpendicular is drawn the vertex of the right angle of a right triangle to the hypotenuse, the triangle on both sides of the perpendicular are similar to the whole triangle and to each other]

Using equation (1)

$△BMD\sim △DMC$

$\frac{BM}{DM}=\frac{MD}{MC}$ [If two triangle are similar, then the ratio of their corresponding sides are equal]

$D{M}^2=BM.MC.....\left(3\right)$

In $△ABC$

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

$\Rightarrow AB||DM$ [Perpendicular drawn to a same line are parallel to each other]

$\therefore NB||DM$

Also,

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

$\Rightarrow CB||DN$ [Perpendicular drawn to a same line are parallel to each other]

$\therefore MB||DN$

Now, in quardrilateral DNBM

$ND||DM$ and $MB||DN$

Since both pairs of opposite sides are parallel

DNMB is a parallelogram

Since opposite sides of parallelogram are equal

$\therefore DN=MB$ and $DM=NB.....\left(4\right)$

Putting $BM=DN$ in equation (3)

$D{M}^2=BM.MC$

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

Hence proved

$\left(ii\right)$ From (2)

$△AND\sim △DNB$

$\frac{AN}{DN}=\frac{DN}{BN}$ [If two triangles are similar, then the ratio of their corresponding sides are equal]

$D{N}^2=AN.BN$

Putting $BN=DM$ from (4)

$D{N}^2=AN.DM$

Hence proved