Triangles : Exercise - 6.6 Optional (Mathematics NCERT Class 10th)

(*These exercises are not for examination point of view)
Q.1     In Figure, PS is the bisector of $\angle QPR$ of $\Delta PQR$. Prove that ${{QS} \over {SR}} ={{PQ} \over {PR}}$ Sol.

Given PQR is a triangle and PS is the internal bisector of $\angle QPR$ meeting QR at S.
Therefore $\angle QPS = \angle SPR$
To prove : ${{QS} \over {SR}} = {{PQ} \over {PR}}$
Construction : Draw RT || SP to cut QP produced at T.
Proof : Since PS || TR and PR cuts them, hence, we have $\angle SPR = \angle PRT$ ............ (1)  [Alternate $\angle s$]
and, $\angle QPS = \angle PTR$ ............... (2)  [Corresponding $\angle s$]
But, $\angle QPS = \angle SPR$ [Given]
Therefore $\angle PRT = \angle PTR$ [From (1) and (2)]
$\Rightarrow$ PT = PR .......... (3) [Because sides opp. to equal $\angle s$ are equal]
Now, in $\Delta QRT$, we have
RT || SP [By construction]
Therefore ${{QS} \over {SR}} = {{PQ} \over {PT}}$ [By Basic Proportionality Theorem]
$\Rightarrow$ ${{QS} \over {SR}} = {{PQ} \over {PR}}$ [From (3)]

Q.2     In figure, D is a point on hypotenuse AC of $\Delta ABC$, $BD \bot AC$, $DM \bot BC$ and $DN \bot AB$. Prove that
(i) $D{M^2} = DN.MC$
(ii) $D{N^2} = DM.AN$ Sol.

We have, $AB \bot BC$ and $DM \bot BC$
$\Rightarrow$ AB || DM
Similarly, we have
$BC \bot AB$ and $DN \bot AB$
$\Rightarrow$ CB || DN Hence, quadrilateral BMDN is a rectangle.
Therefore BM = ND
(i)   In $\Delta BMD$, we have
$\angle 1 + \angle BMD + \angle 2 = 180^\circ$
$\Rightarrow$ $\angle 1 + 90^\circ + \angle 2 = 180^\circ \Rightarrow \angle 1 + \angle 2 = 90^\circ$

Similarly, in $\Delta DMC$, we have
$\angle 3 + \angle 4 = 90^\circ$
Since $BD \bot AC$. Therefore,
$\angle 2 + \angle 3 = 90^\circ$
Now, $\angle 1 + \angle 2 = 90^\circ$ and $\angle 2 + \angle 3 = 90^\circ$
$\Rightarrow$ $\angle 1 + \angle 2 = \angle 2 + \angle 3$
$\Rightarrow$ $\angle 1 = \angle 3$
Also, $\angle 3 + \angle 4 = 90^\circ$ and $\angle 2 + \angle 3 = 90^\circ$
$\Rightarrow$ $\angle 3 + \angle 4 = \angle 2 + \angle 3 \Rightarrow \angle 2 = \angle 4$
Thus, in $\Delta s$ BMD and DMC, we have
$\angle 1 = \angle 3$ and $\angle 2 = \angle 4$
Therefore By AA-criterion of similarity, we have
$\Delta BMD \sim \Delta DMC$
$\Rightarrow$ ${{BM} \over {DM}} = {{MD} \over {MC}}$
$\Rightarrow$ ${{DN} \over {DM}} = {{DM} \over {MC}}$ [Because BM = ND]
$\Rightarrow$ $D{M^2} = DN \times MC$
(ii)   Proceeding as in (i), we can prove that
$\Delta BND \sim \Delta DNA$
$\Rightarrow$ ${{BN} \over {DN}} = {{ND} \over {NA}}$
$\Rightarrow$ ${{DM} \over {DN}} = {{DN} \over {AN}}$ [Because BN = DM]
$\Rightarrow$ $D{N^2} = DM \times AN$

Q.3      In figure, ABC is a triangle in which $\angle ABC > 90^\circ$ and $AD \bot CB$ produced. Prove that $A{C^2} = A{B^2} + B{C^2} + 2BC.BD$ Sol.

Given : ABC is a triangle in which $\angle ABC > 90^\circ$ and $AD \bot CB$ produced.
To Prove : $A{C^2} = A{B^2} + B{C^2} + 2BC.BD$
Proof : Since $\Delta ADB$ is a right triangle, right-angled at D, therefore, by Pythagoras theorem, we have
$A{B^2} = A{D^2} + D{B^2}$ ........... (1) Again, $\Delta ADC$ is a right triangle, right-angled at D. Therefore, by Pythagoras theorem, we have
$A{C^2} = A{D^2} + D{C^2}$
$\Rightarrow$ $A{C^2} = A{D^2} + {\left( {DB + BC} \right)^2}$
$\Rightarrow$ $A{C^2} = A{D^2} + D{B^2} + B{C^2} + 2DB.BC$
$\Rightarrow$ $A{C^2} = \left( {A{D^2} + D{B^2}} \right) + B{C^2} + 2BC.BD$
$\Rightarrow$ $A{C^2} = A{B^2} + B{C^2} + 2BC.BD$ [Using (1)]
which proves the required result.

Q.4      In figures, ABC is a triangle in which $\angle ABC < 90^\circ$ and $AD \bot BC$. Prove that $A{C^2} = A{B^2} + B{C^2} - 2BC.BD$ Sol.

Given : ABC is a triangle in which $\angle ABC < 90^\circ$ and $AD \bot BC$.
To prove : $A{C^2} = A{B^2} + B{C^2} - 2BC.BD$
Proof : Since $\Delta ADB$ is a right triangle, right-angled at D, therefore, by Pythagoras theorem, we have,
$A{B^2} = A{D^2} + B{D^2}$ .............. (1)
Again, $\Delta ADC$ is a right triangle, right-angled at D, therefore, by Pythagoras theorem, we have,
$A{C^2} = A{D^2} + D{C^2}$
$\Rightarrow$ $A{C^2} = A{D^2} + {\left( {BC - BD} \right)^2}$
$\Rightarrow$ $A{C^2} = A{D^2} + \left( {B{C^2} + B{D^2} - 2BC.BD} \right)$
$\Rightarrow$ $A{C^2} = \left( {A{D^2} + B{D^2}} \right) + B{C^2} - 2BC.BD$
$\Rightarrow$ $A{C^2} = A{B^2} + B{C^2} - 2BC.BD$ [Using (1)]
which proves the required result.

Q.5      In figure, AD is a median of a triangle ABC and $AM \bot BC$. Prove that
(i) $A{C^2} = A{D^2} + BC.DM + {\left( {{{BC} \over 2}} \right)^2}$
(ii) $A{B^2} = A{D^2} - BC.DM + {\left( {{{BC} \over 2}} \right)^2}$
(iii) $A{C^2} = A{B^2} = 2A{D^2} + {1 \over 2}B{C^2}$ Sol.

Since $\angle AMD = 90^\circ$, therefore, $\angle ADM < 90^\circ$ and $\angle ADC > 90^\circ$.
Thus, $\angle ADC$ is acute and $\angle ADC$ is obtuse.
(i)    In $\Delta ADC$, $\angle ADC$ is an obtuse angle. Therefore $A{C^2} = A{D^2} + D{C^2} + 2DC.DM$

$\Rightarrow$ $A{C^2} = A{D^2} + {\left( {{{BC} \over 2}} \right)^2} + 2.{{BC} \over 2}.DM$

$\Rightarrow$ $A{C^2} = A{D^2} + {\left( {{{BC} \over 2}} \right)^2} + BC.DM$
$\Rightarrow$ $A{C^2} = A{D^2} + BC.DM + {\left( {{{BC} \over 2}} \right)^2}$ .............. (1)
(ii)     In $\Delta ABD$, $\angle ADM$ is an acute angle.
Therefore $A{B^2} = A{D^2} + B{D^2} - 2BD.DM$
$\Rightarrow$ $A{B^2} = A{D^2} + {\left( {{{BC} \over 2}} \right)^2} - 2.{{BC} \over 2}.DM$

$\Rightarrow$ $A{B^2} = A{D^2} - BC.DM + {\left( {{{BC} \over 2}} \right)^2}$ ............ (2)
(iii)    From (1) and (2), we get
$A{B^2} + A{C^2} = 2A{D^2} + {1 \over 2}B{C^2}$

Q.6      Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Sol.

We know that if AD is a median of $\Delta ABC$, then $A{B^2} + A{C^2} = 2A{D^2} + {1 \over 2}B{C^2}$
[See above question part (iii)]
Sine the diagonals of a parallelogram bisect each other, therefore, BO and DO are medians of $\Delta s$ Therefore $A{B^2} + B{C^2} = 2B{O^2} + {1 \over 2}A{C^2}$ ........... (1)
and, $A{D^2} + C{D^2} = 2D{O^2} + {1 \over 2}A{C^2}$ .......... (2)
Adding (1) and (2), we get
$A{B^2} + B{C^2} + C{D^2} + A{D^2} = 2\left( {B{O^2} + D{O^2}} \right) + A{C^2}$
$\Rightarrow$ $A{B^2} + B{C^2} + C{D^2} + A{D^2} = 2\left( {{1 \over 4}B{D^2} + {1 \over 4} B{D^2}} \right) + A{C^2}$ [Because $DO = {1 \over 2}BD$]
$\Rightarrow$ $A{B^2} + B{C^2} + C{D^2} + A{D^2} = A{C^2} + B{D^2}$

Q.7      In figure, two chords AB and CD intersect each other at the point P. Prove that
(i) $\Delta APC \sim \Delta DPB$
(ii) AP.PB = CP.DP Sol.

(i)   In $\Delta s$ APC and DPB, we have
$\angle APC = \angle DPB$ [Vert. opp. $\angle s$]
$\angle CAP = \angle BDP$ [Angles in the same segment of a circle are equal]
Therefore by AA-criterion of similarity, we have
$\Delta APC \sim \Delta DPB$
(ii)  Since $\Delta APC \sim \Delta DPB$
Therefore ${{AP} \over {DP}} = {{CP} \over {PB}}$
$\Rightarrow$ $AP \times PB = CP \times DP$

Q.8      In figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle, Prove that
(i) $\Delta PAC \sim \Delta PDB$
(ii) PA.PB = PC.PD Sol.

(i)    In $\Delta s$ PAC and PDB, we have
$\angle APC = \angle BPD$ [Common]
$\angle PAC = \angle PDB$ [Because $\angle BAC = 180^\circ - \angle PAC$ and $\angle PDB = \angle CDB = 180^\circ - \angle BAC = 180^\circ - \left( {180^\circ - \angle PAC} \right) = \angle PAC$]

Therefore by AA-criterion of similarity, we have
$\Delta PAC \sim \Delta DPB$
(ii)   Since $\Delta PAC \sim \Delta DPB$
${{PA} \over {PD}} = {{PC} \over {PB}}$
$\Rightarrow$ PA.PB = PC.PD

Q.9     In figure, D is a point on side BC of $\Delta ABC$ such that ${{BD} \over {CD}} = {{AB} \over {AC}}$. Prove that AD is the bisector of $\angle BAC$. Sol.

Given : ABC is a triangle and D is a point on BC such that
${{BD} \over {CD}} = {{AB} \over {AC}}$
To prove : AD is the internal bisector of $\angle BAC$.
Construction : Produce BA to E such that AE = AC. Join CE. Proof : In $\Delta AEC$, since AE = AC, hence
$\angle AEC = \angle ACE$ ........... (1)
[Because Angles opp. to equal sides of a $\Delta$ are equal]
Now, ${{BD} \over {CD}} = {{AB} \over {AC}}$ [Given]
$\Rightarrow$ ${{BD} \over {CD}} = {{AB} \over {AE}}$ [Because AE = AC, construction]
Therefore by converse of Basic Proportionality Theorem, we have
DA || CE
Now, since CA is a transeversal, we have
$\angle BAD = \angle AEC$ ......... (2) [Corresponding $\angle s$]
and, $\angle DAC = \angle ACE$ ...... (3) [Alternate angles]
Also, $\angle AEC = \angle ACE$ [From (1)]
Hence, $\angle BAD = \angle DAC$ [From (2) and (3)]
Thus, AD bisects, $\angle BAC$ internally.

Q.10 Nazima is fly fishing i a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 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 taur, how much string does she have out (see figure) ? If she pulls in the string at the rate of 5 cm per second, what will the horizontal distance of the fly from her after 12 seconds ? Sol.

In fact, we want to find AC.
By Pythagoras theorem, we have
$A{C^2} = {\left( {2.4} \right)^2} + {\left( {1.8} \right)^2}$ $\Rightarrow$ $A{C^2} = 5.76 + 3.24 = 9.00$
$\Rightarrow$ AC = 3 m
Therefore length of string she have out = 3 m.
Length of the string pulled at the rate of 5 cm/sec in 12 seconds.
= (5 × 12) cm
= 60 cm = 0.60 m
Therefore remaining string left out = (3 – 0.6) m = 2.4 m
In 2nd case let us find PB
$P{B^2} = P{C^2} - B{C^2}$
= ${\left( {2.4} \right)^2} - {\left( {1.8} \right)^2}$
= $5.76 - 3.24 = 2.52$
$\Rightarrow$ $PB = \sqrt {2.52} = 1.59$ (nearly) Hence, the horizontal distance of the fly from Nazima after 12 seconds
= (1.59 + 1.2) m
= 2.79 m (nearly)