Ex. 6.6 Q2 Triangles Solution - NCERT Maths Class 10

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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:

\(\begin{align} \left( {\text{a}} \right)\,\,D{M^2} = DN.MC\\ \left( {\text{b}} \right)\,\,\,D{N^2} = DM.AN \end{align}\)

Text Solution

Reasoning:

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

Steps:

(i) In quadrilateral \(DMBN\)    
               \( DM \bot BC\;\) and \(\;DN \bot AB\) 
               \(DMBN\) is a rectangle.
               \(DM = BN\) and \(DN = MB\) ...... (i)

In \(\Delta DCM\)
\[\begin{align} &\angle DCM + \angle DMC + \angle CDM = {180^ \circ }\\& \angle DCM + {90^ \circ } + \angle CDM = {180^ \circ }\\& \angle DCM + \angle CDM = {90^ \circ } .... \rm\left (ii \right)\end{align}\]
But \(\angle CDM + \angle BDM = {90^ \circ }.....\rm{}\left( {iii} \right) \)
Since \(\,BD \bot AC\) 

From (ii) and (iii)
\[\begin{align}\angle D C M+\angle B D M \ldots (\mathrm{iv})\end{align}\]

In \(\Delta BDM\)
                  \(\begin{equation} \begin{array}{l}{\angle D B M+\angle B D M=90^{\circ}}\ldots (\mathrm{v}) \\ {\text{Since}\;,D M \perp B C }\end{array} \end{equation}\)

From (iii) and (v) 

\(\begin{equation} \angle C D M=\angle D B M \ldots \rm{(vi)} \end{equation}\)

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

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

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

(ii) In \(\Delta B D N\)

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

From (vii) and (viii) 

\(\angle D B N=\angle A D N \dots \rm{(ix)} \)

In \(\Delta A D N \)
\[\begin{align}{\angle DAN+\angle ADN} &= {90^{\circ}(\text { Since } D N \perp A B) \ldots (\mathrm{x})} \\ {\mathrm{But}\, \angle BDN+\angle ADN}& = {90^{\circ}(\text { Since } BD \perp A C)}\ldots \rm{(xi)}\end{align}\]

From (xi) and (x) 
            \(\angle D A N=\angle B D N \ldots (\mathrm{xii})\)

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

\(\Delta B D N \sim \Delta \mathrm{DAN}\)            (From (ix) and (xii), \(AA\) criterion) 

\(\begin{align}\frac{B N}{D N}=\frac{D N}{A N}\end{align}\)                   ( Corresponding sides are in same ratio)
\[\begin{align}{D N^{2}}&={B N \cdot A N} \\ {D N^{2}}&={D N \cdot A N\qquad[\text{from (i)B N=D M}]}\end{align}\]

  
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