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In the given figure, $D$ is a point on hypotenuse $AC$ of $∆ABC, DM \perp BC$ and $DN \perp AB$. Prove that:
(i) $DM^2 = DN \times MC$
(ii) $DN^2 = DM \times AN$
"
Given:
$D$ is a point on hypotenuse $AC$ of $∆ABC, DM \perp BC$ and $DN \perp AB$.
To do:
We have to prove that
(i) $DM^2 = DN \times MC$
(ii) $DN^2 = DM \times AN$
(i) In quadrilateral $MBND$,
$\angle \mathrm{M}+\angle \mathrm{B}+\angle \mathrm{N}+\angle \mathrm{D}=360^{\circ}$
$90^{\circ}+90^{\circ}+90^{\circ}+\angle \mathrm{D}=360^{\circ}$
$\angle \mathrm{D}=360^{\circ}-270^{\circ}$
$=90^{\circ}$
This implies,
Quadrilateral $MBND$ is a rectangle.
Therefore,
$\angle \mathrm{NDB}=\angle \mathrm{MDB}=45^{\circ}$ (Diagonals of a rectangle bisect the angles)
$\angle \mathrm{CDM}=\angle \mathrm{ADN}=45^{\circ}$
In $\triangle \mathrm{CMD}$ and $\triangle \mathrm{BMD}$,
$\angle \mathrm{CMD}=\angle \mathrm{BMD}$
$\angle \mathrm{CDM}=\angle \mathrm{BDM}$
Therefore, by AA similarity,
$\triangle \mathrm{CMD} \sim \triangle \mathrm{BMD}$
This implies,
$\frac{\mathrm{DM}}{\mathrm{MB}}=\frac{\mathrm{MC}}{\mathrm{DM}}$ (By BPT)
$\mathrm{DM}^{2}=\mathrm{MB} \times \mathrm{MC}$
$=\mathrm{DN} \times \mathrm{MC}$ (Since $MB = DN$)
Hence proved.
(ii) $\triangle \mathrm{CMD} \sim \triangle \mathrm{BMD}$
This implies,
$\frac{\mathrm{DM}}{\mathrm{MB}}=\frac{\mathrm{MC}}{\mathrm{DM}}$ (By BPT)
$\mathrm{DM}^{2}=\mathrm{MB} \times \mathrm{MC}$
$=\mathrm{DN} \times \mathrm{MC}$ (Since $MB = DN$)
In $\triangle \mathrm{DNB}$ and $\triangle \mathrm{AND}$,
$\angle \mathrm{DMB}=\angle \mathrm{AND}$
$\angle \mathrm{ADN}=\angle \mathrm{BDN}$
Therefore, by AA similarity,
$\triangle \mathrm{DNB} \sim \triangle \mathrm{AND}$
This implies,
$\frac{\mathrm{DN}}{\mathrm{AN}}=\frac{\mathrm{NB}}{\mathrm{DN}}$ (By BPT)
$\mathrm{DN}^{2}=\mathrm{AN} \times \mathrm{NB}$
$\mathrm{DN}^{2}=\mathrm{DM} \times \mathrm{AN}$ (Since $NB = DM$)
Hence proved.
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