Ex. 6.6 Q2 Triangles Solution - NCERT Maths Class 10

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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
Ex 6.6 | Question 2

Text Solution


\(\rm AA\) similarity criterion \( \rm ,BPT.\)


(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 \(\Delta DCM\)

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


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

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 \(\begin{equation} \Delta D C M\;\text{and}\;\Delta D B M \end{equation}\)

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

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

(ii) In \(\Delta B D N\)

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

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

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

In \(\Delta A D N \)

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


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

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

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

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

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

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

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