# Ex. 6.6 Q2 Triangles Solution - NCERT Maths Class 10

Go back to  'Ex.6.6'

## Question

In Fig. below, $$D$$ is a point on hypotenuse $$AC$$ of $$\Delta ABC$$, such that  $$BD \bot AC,\;DM \bot BC{\text{ and }}DN \bot AB.$$

Prove that:

(a) $$\,D{M^2} = DN.MC$$

(b) $$\,D{N^2} = DM.AN$$

Video Solution
Triangles
Ex 6.6 | Question 2

## Text Solution

Reasoning:

$$\rm AA$$ similarity criterion $$\rm ,BPT.$$

Steps:

(i) In quadrilateral $$DMBN$$

$$DM \bot BC\;$$ and $$\;DN \bot AB$$

$$DMBN$$ is a rectangle.

$$DM\!=\!BN \; \text{ and } \; DN\!=\!MB \;\cdots \rm (i)$$

In $$\Delta DCM$$

\begin{align} &\! \! \angle DCM\!+\! \!\angle DMC \! + \!\! \angle CDM \!\!= \!\!180^ \circ \\&\! \! \angle DCM + {90^ \circ } + \angle CDM = {180^ \circ }\\&\! \! \angle DCM + \angle CDM = {90^ \circ } \;\;\dots \rm\left (ii \right)\end{align}

But,

\begin{align} & \angle CDM + \angle BDM = {90^ \circ }\;\;\dots\rm{}\left( {iii} \right)\\ &\qquad \text{Since }\,BD \bot AC \end{align}

From $$\rm(ii)$$ and $$\rm(iii)$$

\begin{align}\angle D C M+\angle B D M \;\; \ldots (\rm{iv})\end{align}

In $$\Delta BDM$$

\begin{align} & \angle D B M+\angle B D M=90^\circ \;\;\dots (\rm{v}) \\& \qquad\text{Since }D M \perp B C \end{align}

From $$\rm (iii)$$ and $$\rm (v)$$

\begin{align} \angle C D M=\angle D B M\;\; \dots \rm{(vi)} \end{align}

Now, in $$$$\Delta D C M\;\text{and}\;\Delta D B M$$$$

\begin{align} & \Delta D C M \sim \Delta D B M \\ & \begin{bmatrix} \text{From (iv) and (vi),}\\ \text{AA Criterion } \end{bmatrix}\\ &\quad \frac{D M}{BM}=\frac{M C}{DM} \\ & \begin{bmatrix} \text{Corresponding sides } \\ \text{are in same ratio}\end{bmatrix} \end{align}

\begin{align} DM^{2}&=B M \cdot M C \\ DM^{2}&=D N \cdot MC \\ [\text{from (i) } &D N=B M]\end{align}

(ii) In $$\Delta B D N$$

\begin{align} & \angle BDN + \angle DBN \!\!= {90^ \circ }\;\dots\left( \rm{vii} \right) \\ & \left( {\rm{Since}\,\,DN \bot AB} \right)\end{align}
But,
\begin{align} & \angle ADN + \angle BDN \!\!= {90^ \circ } \;\dots\left( \rm{viii} \right) \\ & \left( {\rm{Since}\,\,BD \bot AC} \right) \end{align}

From $$\rm(vii)$$ and $$\rm(viii)$$

\begin{align} \angle D B N=\angle A D N \;\dots \rm{(ix)} \end{align}

In $$\Delta A D N$$

\begin{align} &\!\! \angle DAN+\angle ADN = 90^{\circ} \; \dots (\text{x}) \\ & (\text { Since } D N \perp A B)\end{align}

But,

\begin{align}& \angle BDN+\angle ADN = 90^{\circ}\; \ldots \rm{(xi)} \\ & (\text {Since } BD \perp A C)\end{align}

From $$\rm(xi)$$ and $$\rm(x)$$

\begin{align}\angle D A N=\angle B D N \;\ldots (\rm{xii})\end{align}

Now in $$\Delta B D N\;\text{and}\; \Delta D A N$$

\begin{align} & \Delta B D N \sim \Delta DAN \\ & \begin{bmatrix} \text{From (ix) and (xii),}\\ \text{AA Criterion } \end{bmatrix} \\&\quad\frac{B N}{D N}=\frac{D N}{A N} \\ & \begin{bmatrix}\text{Corresponding sides }\\ \text{are in same ratio}\end{bmatrix} \end{align}

\begin{align}{D N^{2}}&={B N \cdot A N} \\ {D N^{2}}&=D N \cdot A N \\ [\text{From (i) }&B N=D M] \end{align}

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school