# NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.6

Go back to  'Triangles'

## Chapter 6 Ex.6.6 Question 1

In below Fig, $$PS$$ is the bisector of $$\angle {{QPR}}$$ of $$\Delta PQR$$. Prove that \begin{align}\frac{{QS}}{{SR}} = \frac{{PQ}}{{PR}}\end{align}

### Solution

Reasoning:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.(BPT)

Steps:

Draw a line parallel to $$PS,$$ through $$R,$$ which intersect $$QP$$ produced at $$T$$

$PS\parallel RT$

In $$\Delta QPR$$

\begin{align}&\angle QPS = \angle SPR \\ &\text{(PS is bisector of }\angle QPR \; \dots (1) \end{align}

But,

\begin{align}& \angle PRT = \angle SPR \\ &\text{(Alternate interior angles)}\; \dots (2) \end{align}

\begin{align}& \angle QPS = \angle PTR \\ &\text{(Corresponding angles)}\; \dots (3) \end{align}

From $$\rm{(1), (2)}$$ and $$\rm{(3)}$$

\begin{align}\angle PTR &= \angle PRT\\ PR &= PT\;.......\rm (4)\end{align}

(Since in a triangle, sides opposite to the equal angles are equal)

In $$\Delta QSP,PS \parallel RT$$

\begin{align}\frac{{QS}}{{SR}}&= \frac{{QP}}{{PT}}\;\;\;\left[ {{\rm{BPT}}} \right]\\\frac{{QS}}{{SR}}&= \frac{{QP}}{{PR}}\;\;\;\left[ {{\rm{from (4)}}} \right]\end{align}

## Chapter 6 Ex.6.6 Question 2

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

### 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}

## Chapter 6 Ex.6.6 Question 3

In Fig. below, $$ABC$$ is a triangle in which $$\Delta ABC > 90°$$ and $$AD \perp CB$$ produced.

Prove that:

$A{C^2} = A{B^2} + B{C^2} + 2BC.BD$

#### ### Solution Video Solution

Reasoning:

Pythagoras theorem

#### Steps:

In $$\Delta ADC$$

\begin{align} \angle ADC &= {90^ \circ }\\ \Rightarrow A{C^2} &= A{D^2} + C{D^2}\\ &= A{D^2} + {\left[ {BD + BC} \right]^2}\\ &=\!\! \begin{bmatrix} A{D^2} + B{D^2} + B{C^2} \\+ 2BC \cdot BD \! \end{bmatrix} \\ A{C^2} &= \begin{bmatrix} A{B^2} + B{C^2} \\ + 2BC \cdot BD \end{bmatrix} \end{align}

Therefore, in

\begin{align}\angle ADB,\;A{B^2} = A{D^2} + B{D^2}\;\end{align}

## Chapter 6 Ex.6.6 Question 4

In Fig below, $$ABC$$ is a triangle in which $$ABC < 90°$$ and $$AD \bot BC.$$
Prove that:

$$A{C^2} = A{B^2} + B{C^2} - 2BC \cdot BD$$

#### ### Solution Video Solution

Reasoning:

Pythagoras Theorem.

#### Steps:

In $$\Delta ADC$$

\begin{align} \angle ADC &= {90^ \circ }\\ A{C^2}& = A{D^2} + D{C^2}\\ &= A{D^2} + {\left[ {BD - BC} \right]^2}\\ &= \begin{bmatrix} A{D^2} + B{D^2} + \\ B{C^2} - 2BC.BD \end{bmatrix} \\ A{C^2}&= \!A{B^2} + B{C^2}\! - \!\!2BC.BD \end{align}

In $$\Delta ADB,$$

\begin{align} A{B^2} = A{D^2} + B{D^2} \end{align}

## Chapter 6 Ex.6.6 Question 5

In, $$AD$$ is a median of a triangle $$ABC$$ and $$AM \bot BC.$$

Prove that:

i) \begin{align} A{C^2} = A{D^2} + BC.DM + {\left[ {\frac{{BC}}{2}} \right]^2} \end{align}

ii) \begin{align} A{B^2} = A{D^2} - BC.DM + {\left[ {\frac{{BC}}{2}} \right]^2} \end{align}

iii) \begin{align} A{C^2} + A{B^2} = 2A{D^2} + \frac{1}{2}B{C^2} \end{align}

#### Reasoning:

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

#### Steps:

$$\rm i)$$ In $$\Delta AMC$$

$$\angle AMC = {90^ \circ }$$

\begin{align} A{C^ 2} &\!=\!A{M^2}\!+\! M{C^2}\\ &\!= \!\! A{M^2} \!+\! {\left[ {MD \!+\! CD} \right]^2} \\ &\!=\! \!A{M^2} \!+\! M{D^2} \!+\! C{D^2} +\! 2MD.CD \\ &\!=\! \!A{D^2} \!+\! {\left[ {\frac{{BC}}{2}} \right]^2} \! +\! 2MD.\left[ {\frac{{BC}}{2}} \right]\end{align}

Since, in $$\Delta AMD,{\rm{ }}A{D^2} = A{M^2} + D{M^2}$$, and  is the midpoint of  means

\begin{align}BD = CD = \frac{{BC}}{2} \end{align}

\begin{align} A{C^2}\! = \! A{D^2} \!+\! MD.BC \!+\! {\left[ {\frac{{BC}}{2}} \right]^2} \;\dots\rm (i) \end{align}

$$\rm ii)$$ In $$\Delta \,AMB$$

$$\angle AMB = {90^ \circ }$$

\begin{align}A{B^2} &\!=\!A{M^2}\! +\! B{M^2}\\ &\!=\! A{M^2}\! +\! {\left[ {BD \!-\! DM} \right]^2}\\ &\!=\! A{M^2}\! +\! B{D^2}\! +\! D{M^2} \! - \!2BD.DM \\ &\!= \! \begin{Bmatrix}A{M^2}\! +\! D{M^2} \!+\! {\left[ {\frac{{BC}}{2}} \right]^2} \!\\-\! 2 { \left[{ \frac{{BC}}{2} }\right]} DM \end{Bmatrix} \end{align}

Since,

In $$\Delta AMD,$$

$$AD^2 = AM^2 + DM^2 \text{and}\;D$$

is the midpoint of $$BC$$ means

$$BD = CD = \frac{{BC}}{2}$$

\begin{align} A{B^2} = \! A{D^2} \! - \! BC.DM \! + \!\! {\left( {\frac{{BC}}{2}} \right)^2}....\left(\rm {ii} \right)\end{align}

$$\rm iii)$$ Adding ($$\rm{i}$$) and ($$\rm{ii}$$)

\begin{align} A{C^2}\!+\!A{B^2} &\!=\!\begin{Bmatrix}[ A{D^2} + {\left[ {\frac{{BC}}{2}} \right]^2} + BC.DM + \\A{D^2}+ {\left[ {\frac{{BC}}{2}} \right]^2} - BC.DM \end{Bmatrix} A{C^2} + A{B^2} \\&= 2A{D^2} + 2{\left[ {\frac{{BC}}{2}} \right]^2}A{C^2} + A{B^2} \\&= 2A{D^2} + \frac{1}{2}{B{C^2}} \end{align}

## Chapter 6 Ex.6.6 Question 6

Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

### Solution

Reasoning:

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

#### Steps:

In parallelogram $$ABCD$$

\begin{align} AB = CD \\ AD = BC\end{align}

Draw $$A E \perp C D,$$ $$D F \perp A B$$

\begin{align} ​​​​& \qquad \qquad EA = DF \\& \begin{bmatrix} \text{Perpendiculars drawn between }\\ \text{same parallel lines}\end{bmatrix} \end{align}

In $$\Delta AEC$$

\begin{align}A{C^2} &\! = \! A{E^2} + E{C^2}\\ &\! =\!  A{E^2} + {\left[ {ED + DC} \right]^2}\\ &\! = \! A{E^2} \! \!+\!\! D{E^2} \!\!+\! \! D{C^2} \!+\! 2DE.\! DC \\ A{C^2} &\! = \! A{D^2}\! + \! D{C^2} \! + \! 2DE \cdot \! DC  \; \ \ldots \left(\text{i} \right) \\ & \left[ {{\text{Since, }}A{D^2} \! =\!  A{E^2} + D{E^2}} \right]\end{align}

In $$\Delta \,DFB$$

\begin{align} B{D^2} &\!=\! D{F^2}\! +\! B{F^2}\\ &\!=\! D{F^2}\! +\! {\left[ {AB\! -\! AF} \right]^2}\\ &\!= \! D{F^2} \!\! +\!\! A{B^2} \!\! +\!\! A{F^2}\! \!-\!\! 2AB. \!AF \\ &\!=\! A{D^2} \!+ \!A{B^2}\! - \!2AB.AF\\ B{D^2} &\!= \!A{D^2} \!+\! A{B^2}\! -\! 2AB.AF \; \ldots \left( \rm{ii} \right) \\ &\text{(Since } A{D^2} = D{F^2} + A{F^2}) \end{align}

\begin{align} A{C^2}\!+\!B{D^2} &\!=\! \begin{bmatrix}A{D^2} \!+\! D{C^2} \! +\! 2DE \cdot DC \!+\! \\ A{D^2} \!+\! A{B^2} \! -\! 2AB.AF \end{bmatrix} \\ A{C^2} \!+\! B{D^2} &\!= \!\begin{bmatrix}B{C^2}\! +\! D{C^2} \!+\! A{D^2} \!+\! A{B^2} \!+\!\\ 2AB.AF \!- \!2AB.F\end{bmatrix} \\ \text{(Since } &AD \!=\! BC, \;DE \!=\! AF,\; CD \!=\! AB) \end{align}

\begin{align} \Rightarrow \quad& A{C^2} + B{D^2} \\ = & A{B^2} \!+\! B{C^2} \!+\! C{D^2} \!+ \! A{D^2} \end{align}

## Chapter 6 Ex.6.6 Question 7

In Fig. below, two chords $$AB$$ and $$CD$$ intersect each other at the point $$P.$$

Prove that:

(i) $$\Delta APC\,\text{~}\Delta {\text{ }}DPB$$

(ii) $$AP. PB = CP. DP$$

### Solution

Reasoning:

As we know that, two triangles, are similar if:

(i) Their corresponding angles are equal.

(ii) Their corresponding sides are in the same ratio.

As we know that angles in the same segment of a circle are equal.

Steps:

(i) In, $$\Delta APC \;\text{and}\; \Delta DPB$$

$$\angle APC = \angle DPB$$

(Vertically opposite angles)

$$\angle PAC = \angle PDB$$

(Angles in the same segment)

$$\Rightarrow \Delta APC\text{~}\Delta DPB$$

(A.A criterion)

(ii) In, $$\Delta APC \;\text{and}\; \Delta DPB$$

\begin{align} \frac{{AP}}{{PD}}& = \frac{{PC}}{{PB}} = \frac{{AC}}{{DB}} \\ & \quad \left[ {\Delta APC\text{~}\Delta DPB} \right]\\ \frac{{AP}}{{PD}} &= \frac{{PC}}{{PB}} \end{align}

$$\Rightarrow \;\;AP.PB = PC.PD$$

## Chapter 6 Ex.6.6 Question 8

In Fig. below, two chords $$AB$$ and $$CD$$ of a circle intersect each other at the point $$P$$ (when produced) outside the circle.

Prove that:

(i) $$\Delta {\text{ }}PAC{\text{ }}\text{~}{\text{ }}\Delta {\text{ }}PDB$$

(ii) $$PA. PB = PC. PD$$

### Solution

Reasoning:

(i) Exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

Steps:

Draw $$AC$$

(i) In, $$\Delta PAC,\,\,\Delta PDB$$

\begin{align} & \angle APC=\angle BPD\,\,\, \\ & (\text{Common}\,\text{angle}) \\ & \\ & \angle PAC=\angle PDB\,\, \\ & \left( \begin{array} & \text{Exterior angle of a cyclic} \\ \,\text{quadrilateral is equal to the} \\ \text{interior opposite angle} \\ \end{array} \right)\, \\ & \\ & \Rightarrow \Delta PAC~\sim \Delta \text{PDB}\, \\ \end{align}

(ii) In, $$\Delta PAC,\,\,\Delta PDB$$

\begin{align} &\frac{{PA}}{{PD}} = \frac{{PC}}{{PB}} = \frac{{AC}}{{BD}}\\ &\frac{{PA}}{{PD}} = \frac{{PC}}{{PB}}\\ &PA \cdot PB = PC \cdot PD \end{align}

## Chapter 6 Ex.6.6 Question 9

In Fig.below, $$D$$ is a point on side $$BC$$ of $$\Delta ABC$$ such that \begin{align}\frac{{BD}}{{CD}} = \frac{{BA}}{{CA}}\end{align}. Prove that $$AD$$ is the bisector of $$\angle BAC$$.

### Solution

#### Reasoning:

(i) As we know that in an isosceles triangle, the angles opposite to equal sides are equal.

(ii) Converse of BPT

#### Steps:

Extended $$BA$$ to $$E$$ such that $$AE = AC$$ and join $$CE.$$

In $$\Delta AEC$$

\begin{align}AE &= AC \\ \Rightarrow \angle ACE &= \angle AEC \;\;\dots\left(\text{i} \right) \end{align}

It is given that

\begin{align} \frac{{BD}}{{CD}} & = \frac{{BA}}{{CA}}\\ \frac{{BD}}{{CD}} & = \frac{{BA}}{{AE}} \;\;\cdots \left( \text{ii} \right) \\ & \left(\because {AC = AE} \right) \end{align}

In $$\Delta ABD$$ and $$\Delta EBC$$

\begin{align} &AD||EC\,\,\left( {{\text{Converse of BPT}}} \right)\\\\ & \Rightarrow \angle BAD = \angle BEC \cdots \left( \text{iii} \right) \\ &\quad \left( \text{Corresponding angles} \right) \\ & \qquad \qquad\text{and} \, \\&\quad\angle DAC = \angle ACE \cdots \left( \text{iv} \right) \\ &\quad \left( \text{Alternative Angles} \right) \end{align}

From $$\rm(i)$$, $$\rm(iii)$$ and $$\rm (iv)$$

\begin{align} &\angle BAD = \angle DAC\\ & \Rightarrow AD\,\text{is the bisector of}\;\angle BAC \end{align}

## Chapter 6 Ex.6.6 Question 10

Nazima is fly fishing in a stream. The tip of her fishing rod is $$1.8\,\rm m$$ above the surface of the water and the fly at the end of the string rests on the water $$3.6\,\rm m$$ away and $$2.4\,\rm m$$ from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. )? If she pulls in the string at the rate of $$5\,\rm cm$$ per second, what will be the horizontal distance of the fly from her after $$12\,\rm seconds$$?

#### Reasoning:

Pythagoras Theorem

#### Steps:

To find $$AB$$ and $$ED$$

$$BD = 3.6\, {\rm m} ,\\ BC = 2.4 \,{\rm m} , \\ CD = 1.2\,{\rm m},\, \\ AC = 1.8\,\rm cm$$

In $$\Delta ACB$$

\begin{align} A{B^2} &= A{C^2} + B{C^2}\\ &= {\left( {1.8} \right)^2} + {\left( {2.4} \right)^2}\\ &= 3.24 + 5.76\\ A{B^2} &= 9.00\\AB &= 3 \end{align}

Length of the string out $$AB = \rm{3\,cm}$$

Let the fly at $$E$$ after $$12\,\rm{seconds}$$

String pulled in $$12\,\rm{ seconds}$$

\begin{align} &= 12 \times 5\\&= 60 \rm{cm}\\ &= 0.6 \rm{m}\\ AE &= 3\,\rm{m}-0.6\, \rm{m}\\ &= 2.4 \,\rm{m} \end{align}

Now in $$\Delta ACE$$

\begin{align} C{E^2}& = A{E^2} - A{C^2}\\ &= {\left( {2.4} \right)^2} - {\left( {1.8} \right)^2}\\ C{E^2} &= 5.76 - 3.24\\ &= 2.52\\ CE& = 1.587\;\rm{}m \end{align}

\begin{align} DE &= CE + CD\\ &= 1.587 + 1.2\\ &= 2.787\\ DE& = 2.79\;\rm{}m \end{align}

Horizontal distance of the fly after $$12\,\rm seconds$$ $$= 2.79\, \rm m$$