In the figure,$D$ is a point on the hypotenuse $AC$ of $\Delta ABC,$ such that $BD \perp AC,$ $DM \perp BC,$ and $DN \perp AB.$ Prove that $DM^{2} = DN \cdot MC.$
(N/A) Join $DB.$
We have $DN \parallel CB,$ $DM \parallel AB,$ and $\angle B = 90^{\circ}.$
Therefore,$DMBN$ is a rectangle.
Thus,$DN = MB$ and $DM = NB.$
Since $D$ is the foot of the perpendicular drawn from $B$ to $AC,$ we have $\angle CDB = 90^{\circ}.$
$\angle 2 + \angle 3 = 90^{\circ} \dots(1)$
In $\Delta CDM, \angle 1 + \angle 2 + \angle DMC = 180^{\circ} \Rightarrow \angle 1 + \angle 2 = 90^{\circ} \dots(2)$
In $\Delta DMB, \angle 3 + \angle DMB + \angle 4 = 180^{\circ} \Rightarrow \angle 3 + \angle 4 = 90^{\circ} \dots(3)$
From $(1)$ and $(2),$ we get $\angle 1 = \angle 3.$
From $(1)$ and $(3),$ we get $\angle 2 = \angle 4.$
In $\Delta DCM$ and $\Delta BDM,$
$\angle 1 = \angle 3$ (Proved above)
$\angle 2 = \angle 4$ (Proved above)
Therefore,$\Delta DCM \sim \Delta BDM$ ($AA$ similarity criterion).
Thus,$\frac{BM}{DM} = \frac{DM}{MC}.$
Since $BM = DN,$ we have $\frac{DN}{DM} = \frac{DM}{MC}.$
Therefore,$DM^{2} = DN \cdot MC.$
We have $DN \parallel CB,$ $DM \parallel AB,$ and $\angle B = 90^{\circ}.$
Therefore,$DMBN$ is a rectangle.
Thus,$DN = MB$ and $DM = NB.$
Since $D$ is the foot of the perpendicular drawn from $B$ to $AC,$ we have $\angle CDB = 90^{\circ}.$
$\angle 2 + \angle 3 = 90^{\circ} \dots(1)$
In $\Delta CDM, \angle 1 + \angle 2 + \angle DMC = 180^{\circ} \Rightarrow \angle 1 + \angle 2 = 90^{\circ} \dots(2)$
In $\Delta DMB, \angle 3 + \angle DMB + \angle 4 = 180^{\circ} \Rightarrow \angle 3 + \angle 4 = 90^{\circ} \dots(3)$
From $(1)$ and $(2),$ we get $\angle 1 = \angle 3.$
From $(1)$ and $(3),$ we get $\angle 2 = \angle 4.$
In $\Delta DCM$ and $\Delta BDM,$
$\angle 1 = \angle 3$ (Proved above)
$\angle 2 = \angle 4$ (Proved above)
Therefore,$\Delta DCM \sim \Delta BDM$ ($AA$ similarity criterion).
Thus,$\frac{BM}{DM} = \frac{DM}{MC}.$
Since $BM = DN,$ we have $\frac{DN}{DM} = \frac{DM}{MC}.$
Therefore,$DM^{2} = DN \cdot MC.$
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Similar Questions
For the correspondence $ABC \leftrightarrow XYZ$ being a similarity, which of the following correctly matches the information in Part $I$ and Part $II$?
Part $I$ Part $II$ $1.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ}$ and $\angle B \cong \angle Y$ $a. SSS$ condition $2.$ In $\Delta ABC$ and $\Delta XYZ, \angle A \cong \angle X, \angle B \cong \angle Y$ and $\angle C \cong \angle Z$ $b. SAS$ condition $3.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ} = \frac{CA}{ZX}$ $c. AAA$ condition - $d. \text{None of the conditions apply}$
| Part $I$ | Part $II$ |
|---|---|
| $1.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ}$ and $\angle B \cong \angle Y$ | $a. SSS$ condition |
| $2.$ In $\Delta ABC$ and $\Delta XYZ, \angle A \cong \angle X, \angle B \cong \angle Y$ and $\angle C \cong \angle Z$ | $b. SAS$ condition |
| $3.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ} = \frac{CA}{ZX}$ | $c. AAA$ condition |
| - | $d. \text{None of the conditions apply}$ |
Medium
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Medium
View SolutionIn $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AC = 25$ and $AM = 16$,then $BM = \dots$
Easy
View SolutionIn trapezium $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. Prove that the correspondence $MAB \leftrightarrow MCD$ between $\Delta MAB$ and $\Delta MCD$ is a similarity.
Medium
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Easy
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