Properties of Triangle

Option (A):

$ \begin{aligned} & 2\left(a^2-b^2\right)=c^2 \quad\quad(1)\\\\ & \lambda=\frac{\sin (x-y)}{\sin z} \quad\quad(2)\\\\ & \cos (n \pi \lambda)=0 \quad\quad(3)\\\\ & \Rightarrow n \lambda=\frac{(2 m+1)}{2}\quad\quad(4) \end{aligned} $

From Eq. (2), we have

$ \lambda=\frac{\sin x(0) y-\cos x \sin y}{\sin z} $

$ \begin{aligned} & \Rightarrow \lambda=\frac{a \cos y-b \cos x}{c} \\\\ & \Rightarrow \lambda=\frac{a\left(\frac{a^2+c^2-b^2}{2 a c}\right)-b\left(\frac{b^2+c^2-a^2}{2 b c}\right)}{2 c} \quad \text { (by sine formula) } \\\\ & \Rightarrow \lambda=\frac{2\left(a^2-b^2\right)}{2 c^2}=\frac{1}{2}\quad\quad(5) \end{aligned} $

Therefore, from Eqs. (4) and (5), we get

$ \frac{x}{2}=\frac{2 m+1}{2} \Rightarrow x=(2 m+1) $

Hence, (A)→(P), (R), (S).

Option (B):


$ \begin{aligned} & 1+\cos 2 x-2 \cos 2 y=2 \sin x \sin y \\\\ \Rightarrow & 2 \cos ^2 x-2\left(2 \cos ^2 y-1\right)=2 \sin x \sin y \\\\ \Rightarrow & 2 \cos ^2 x-4 \cos ^2 y+2=2 \sin x \sin y \\\\ \Rightarrow & 2 \sin ^2 y-2 \sin x \sin y+\sin x \sin y-\sin ^2 x=0 \\\\ \Rightarrow & 2 \sin y(\sin y-\sin x)+\sin x(\sin y-\sin x)=0 \\\\ \Rightarrow & (\sin y-\sin x)(2 \sin y+\sin x)=0 \\\\ \Rightarrow & b=a \text { or } 2 b=-a(\text { which is impossible }) \\\\ \Rightarrow & \frac{a}{b}=1 \end{aligned} $

Hence, $(\mathrm{B}) \rightarrow(\mathrm{P})$

Option (C):


Vector along the bisector of acute angle between $\overrightarrow{O X}$ and $\overrightarrow{O Y}$ is

$ \frac{\sqrt{3} \hat{i}+\hat{j}}{2}+\frac{\hat{i}+\sqrt{3} \hat{j}}{2}=\frac{(\sqrt{3}+1)}{2}(\hat{i}+\hat{j}) $

Slope of $\overrightarrow{O B}=\tan \left(\frac{\pi}{4}\right)=1$

Equation of $O B$ is $y=x$.

Since

$ \begin{aligned} Z L & =\frac{3}{\sqrt{3}} \Rightarrow \frac{|\beta-(1-\beta)|}{\sqrt{2}}=\frac{3}{\sqrt{2}} \\\\ & \Rightarrow|2 \beta-1|=3 \\\\ & \Rightarrow(2 \beta-1)= \pm 3 \\\\ & \Rightarrow \beta=2 \text { or } \beta=-1 \\\\ & \Rightarrow|\beta|=1 \text { or } 2 \end{aligned} $

$(\mathrm{C}) \rightarrow(\mathrm{P}),(\mathrm{Q})$

Option (D):


$ \Rightarrow y=|\alpha x-1|+|\alpha x-2|+\alpha x ; \alpha \in\{0,1\} $

Case (I): For $\alpha=0, y=3$

Case (II): For $\alpha=1, y=|x-1|+|x-2|+x$

$\Rightarrow y=\left\{\begin{array}{cc}3-x ; & x \leq 1 \\ x+1 ; & 1 < x < 2 \\ 3 x-3 ; & x \geq 2\end{array}\right.$

Hence,

$ \begin{aligned} F(0) & =\int_0^2(3-2 \sqrt{x}) d x=\left[3 x-\frac{4}{3} x^{3 / 2}\right]_0^2 \\\\ & =\left[6-\frac{4}{3}(2 \sqrt{2})\right]=6-\frac{8}{3} \sqrt{2} \\\\ \Rightarrow & F(0)+\frac{8}{3} \sqrt{2}=6 \Rightarrow(T) \end{aligned} $

and

$ \begin{aligned} F(1) & =F(0)-\text { area of } \triangle A C D \\\\ & =\left(6-\frac{8}{3} \sqrt{2}\right)-\frac{1}{2}(2)(1)=5-\frac{8}{3} \sqrt{2} \\\\ & \Rightarrow F(1)+\frac{8}{3} \sqrt{2}=5 \Rightarrow(S) \end{aligned} $

Hence, $(\mathrm{D}) \rightarrow(\mathrm{T}),(\mathrm{S})$

Let $a, b, c$ are the sides of a triangle

And $\quad a+b=x, a b =y$

Given, $\quad x^2-c^2 =y$

$\begin{array}{lrl} \Rightarrow & (x+c)(x-c) & =y \\ \Rightarrow & (a+b+c)(a+b-c) & =y \\ \Rightarrow & 2 s(2 s-2 c) & =y \\ \Rightarrow & s(s-c) & =\frac{y}{4} \quad \text{... (i)}\\ \Rightarrow & \frac{s(s-c)}{y} & =\frac{1}{4} \end{array}$

$\begin{array}{ll} \Rightarrow & \frac{s(s-c)}{a b}=\frac{1}{4} \\ \Rightarrow & \cos ^2 \frac{C}{2}=\frac{1}{4} \\ \Rightarrow & \cos \frac{C}{2}= \pm \frac{1}{2} \\ \Rightarrow & \frac{C}{2}=\frac{\pi}{3}, \frac{2 \pi}{3} \\ \Rightarrow & \mathrm{C}=\frac{2 \pi}{3}, \frac{4 \pi}{3} \end{array}$

$\because \quad C=\frac{4 \pi}{3}$ is impossible in a triangle

$\therefore \quad C=\frac{2 \pi}{3}$

Now, $\frac{r}{\mathrm{R}}=\frac{(s-c) \tan \frac{\mathrm{C}}{2}}{\frac{c}{2 \sin \mathrm{C}}}$

$\begin{aligned} \Rightarrow \quad \frac{r}{\mathrm{R}} & =\frac{(s-c) \cdot \tan \frac{\pi}{3}}{\frac{c}{2 \sin \frac{2 \pi}{3}}} \\ \Rightarrow \quad \frac{r}{\mathrm{R}} & =\frac{\frac{y}{4 s} \cdot \sqrt{3}}{\frac{c}{2 \cdot \frac{\sqrt{3}}{2}}} \quad \left[(s-c)=\frac{y}{4 s} \text { from equation (i) }\right] \end{aligned}$

$\begin{aligned} & \Rightarrow \quad \frac{r}{\mathrm{R}}=\frac{3 y}{2 \cdot 2 s \cdot c} \\ & \Rightarrow \quad \frac{r}{\mathrm{R}}=\frac{3 y}{2(a+b+c) c} \\ & \Rightarrow \quad \frac{r}{\mathrm{R}}=\frac{3 y}{2(x+c) \cdot c} \end{aligned}$

Hint:

$\Rightarrow$ In a triangle $\triangle \mathrm{ABC}$

(i) $r=(\mathrm{s}-c) \tan \frac{\mathrm{C}}{2}, \mathrm{R}=\frac{c}{2 \sin \mathrm{C}}$

(ii) $\cos \frac{c}{2}=\sqrt{\frac{s(s-c)}{a b}}, s=\frac{a+b+c}{2}$

Given, $\Delta$ be the area of $\triangle \mathrm{PQR}$ of side length

$\begin{gathered} a=2, b=\frac{7}{2} \text { and } c=\frac{5}{2} \\ \Rightarrow s=\frac{a+b+c}{2}=\frac{2+\frac{7}{2}+\frac{5}{2}}{2}=4 \end{gathered}$

$\begin{aligned} \text { Now, } & \frac{2 \sin \mathrm{P}-\sin 2 \mathrm{P}}{2 \sin \mathrm{P}+\sin 2 \mathrm{P}} \\ &=\frac{2 \sin \mathrm{P}-2 \sin \mathrm{P} \cdot \cos \mathrm{P}}{2 \sin \mathrm{P}+2 \sin \mathrm{P} \cdot \cos \mathrm{P}} \\ & \quad(\sin 2 \theta=2 \sin \theta \cdot \cos \theta) \end{aligned}$

$\begin{aligned} & =\frac{1-\cos \mathrm{P}}{1+\cos \mathrm{P}} \\ & =\frac{1-\left(1-2 \sin ^2 \frac{\mathrm{P}}{2}\right)}{1+\left(2 \cos ^2 \frac{\mathrm{P}}{2}-1\right)} \\ & \quad\left(\cos 2 \theta=2 \cos ^2 \theta-1=1-2 \sin ^2 \theta\right) \\ & =\tan ^2 \frac{\mathrm{P}}{2} \\ & =\left(\sqrt{\left.\frac{(s-b)(s-c)}{s(s-a)}\right)^2}\right. \\ & =\left(\frac{(s-b)(s-c)}{\sqrt{s(s-a)(s-b)(s-c)}}\right)^2 \end{aligned}$

$\begin{aligned} & =\left(\frac{(s-b)(s-c)}{\Delta}\right)^2 \\ & \therefore \quad(\Delta=\sqrt{s(s-a)(s-b)(s-c)}) \\ & =\frac{\left(4-\frac{7}{2}\right)^2\left(4-\frac{5}{2}\right)^2}{\Delta^2} \\ & =\frac{9}{16 \Delta^2} \\ & =\left(\frac{3}{4 \Delta}\right)^2 \end{aligned}$

Using, $\cos C = {{{a^2} + {b^2} - {c^2}} \over {2ab}}$

$ \Rightarrow {{\sqrt 3 } \over 2} = {{{{({x^2} + x + 1)}^2} + {{({x^2} - 1)}^2} - {{(2x + 1)}^2}} \over {2({x^2} + x + 1)({x^2} - 1)}}$

$ \Rightarrow (x + 2)(x + 1)(x - 1)x + {({x^2} - 1)^2} = \sqrt 3 ({x^2} + x + 1)({x^2} - 1)$

$ \Rightarrow {x^2} + 2x + ({x^2} - 1) = \sqrt 3 ({x^2} + x + 1)$

$ \Rightarrow (2 - \sqrt 3 ){x^2} + (2 - \sqrt 3 )x - (\sqrt 3 + 1) = 0$

$ \Rightarrow x = - (2 + \sqrt 3 )$ and $x = 1 + \sqrt 3 $

But, $x = - (2 + \sqrt 3 ) \Rightarrow c$ is negative.

$\therefore$ $x = 1 + \sqrt 3 $ is the only solution.

Hence, (b) is the correct option.

Since, A, B, C are in AP

$\Rightarrow$ 2B = A + C i.e., $\angle$B = 60$^\circ$

$\therefore$ ${a \over c}$(2 sin C cos C) + ${c \over a}$ (2 sin A cos A)

= 2k (a cos C + c cos A)

[using, ${a \over {\sin A}} = {b \over {\sin B}} = {c \over {\sin C}} = {1 \over k}$]

= 2k (b)

= 2 sin B

[using, b = a cos C + c cos A]

= $\sqrt3$

$ \begin{aligned} \text { Given, } & \mathrm{AB} & =\mathrm{AC} \\ \text { and } & \angle \mathrm{A} & =120^{\circ} \end{aligned} $

Let, $\quad \mathrm{AB}=\mathrm{AC}=b$

and $\quad \mathrm{BC}=a$

Const: Draw $\mathrm{AD} \perp \mathrm{BC}$

Since, AD is also bisector of unequal side BC .

Therefore, area of $\triangle \mathrm{ABC}$

$ \begin{aligned} & =\frac{1}{2} \times \mathrm{BC} \times \mathrm{AD} \\ & =\frac{a}{2}(\mathrm{AD}) \end{aligned} $

To calculate AD, we use trigonometry In $\Delta \mathrm{ADB}$

$ \begin{aligned} \tan 30^{\circ} & =\frac{\mathrm{AD}}{\mathrm{BD}} \\ \Rightarrow \quad \frac{1}{\sqrt{3}} & =\frac{\mathrm{AD}}{\frac{a}{2}} \\ \Rightarrow \quad \mathrm{AD} & =\frac{a}{2 \sqrt{3}} \\ \text { Area of } \triangle \mathrm{ABC} & =\frac{a^2}{4 \sqrt{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

We know that the relation.

Area of triangle $=$ semi-perimeter $\times$ in radius

Now, area of $\Delta \mathrm{ABC}=\left(\frac{a+2 b}{2}\right) \times \sqrt{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \begin{aligned} &\text { From (i) and (ii) we have. }\\ &\left(\frac{a+2 b}{2}\right) \sqrt{3}=\frac{a^2}{4 \sqrt{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

Also in right $\triangle \mathrm{ABD}$

$ \begin{aligned} & & \cos 30^{\circ} & =\frac{\mathrm{BD}}{\mathrm{AB}} \\ & & \frac{\sqrt{3}}{2} & =\frac{a}{2} \\ \Rightarrow & & a & =\sqrt{3} b \end{aligned} $

Substitute $a=\sqrt{3} b$ in (iii), we have

$ \begin{aligned} & \left(\frac{\sqrt{3} b+2 b}{2}\right) \sqrt{3}=\frac{(\sqrt{3} b)^2}{4 \sqrt{3}} \\ & \Rightarrow \quad 3 b+2 \sqrt{3} b=\frac{3 b^2}{2 \sqrt{3}} \\ & \Rightarrow \quad b=\frac{6 \sqrt{3}+12}{3} \\ & \Rightarrow \quad b=(2 \sqrt{3}+4) \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, a=\sqrt{3}(2 \sqrt{3}+4)=6+4 \sqrt{3} \\ & \text { Area, }=\frac{1}{2}\left(\frac{6+4 \sqrt{3}}{2}\right)(2 \sqrt{3}+4) \\ & =(3+2 \sqrt{3})(\sqrt{3}+2) \\ & =3 \sqrt{3}+6+6+4 \sqrt{3}=12+7 \sqrt{3} \end{aligned} $

For option (a) $\cos P = {{{q^2} + {r^2} - {p^2}} \over {2qr}}$ ..... (i)


$\because$ ${{{q^2} + {r^2}} \over 2} \ge \sqrt {{q^2}{r^2}} (AM \ge GM)$

$ \Rightarrow {q^2} + {r^2} \ge 2qr$

From Eq. (i), we get

$\cos P \ge {{2qr - {p^2}} \over {2qr}}$

$ \Rightarrow \cos P \ge 1 - {{{p^2}} \over {2qr}}$ $\to$ option (a) is correct.

For option (b)

${{(q - r)\cos P + (p - r)\cos Q} \over {p + q}}$$ = {{(q\cos P + p\cos Q) - r(\cos P + \cos Q)} \over {p + q}}$

$ \Rightarrow 4{\lambda ^2} - 2\lambda = 0$

$ \Rightarrow u\,.\,v\, = 0$ (Rejected)

or $u\,.\,v = {1 \over 2}$

$\therefore$ $u\,.\,v = {1 \over 2}$

$\therefore$ ${\left| {3u + 5v} \right|^2} = 9{\left| u \right|^2} + 25{\left| v \right|^2} + 3 \times 5 \times 2 \times u\,.\,v$

$ = 9 + 25 + 30 \times {1 \over 2}$

$ = 49$ ($\because$ $\left| u \right| = \left| v \right| = 1$, given)

$\therefore$ $\left| {3u + 5v} \right| = 7$

$ = {{r - r(\cos P + \cos Q)} \over {p + q}} = {{r - r\cos P - r\cos Q} \over {p + q}}$

$ = {{r - (q - \cos R) - (p - q\cos R)} \over {p + q}}$

$ = {{(r - p - q) + (p + q)\cos R} \over {p + q}}$

$ = \cos R + {{r - (p + q)} \over {p + q}} \le \cos R$ ($\because$ $r < p + q$)

Hence, option (b) is correct.

For option (c)

${{q + r} \over p} = {{\sin Q + \sin R} \over {\sin P}} \ge {{2\sqrt {\sin Q.\sin R} } \over {\sin P}}$

$\to$ option (c) is incorrect.

For option (d)

If p < q and p < r, then p is the smallest side and hence one of Q or R can be obtuse.

So, one of cos Q and cos R can be negative.

Therefore, $\cos Q > {p \over r}$ and $\cos R > {p \over q}$ cannot hold always.

Option (d) is incorrect.

For a $\Delta $XYZ, it is given that

$\tan {X \over 2} + \tan {Z \over 2} = {{2y} \over {x + y + z}}$

$ \Rightarrow {\Delta \over {s(s - x)}} + {\Delta \over {s(s - z)}} = {y \over s}$

$ \Rightarrow \Delta {{(s - z + s - x)} \over {(s - x)(s - z)}} = y$

$ \Rightarrow \Delta = (s - x)(s - z)$

$ \Rightarrow s(s - x)(s - y)(s - z) = {(s - x)^2}{(s - z)^2}$

$ \Rightarrow {s^2} - sy = {s^2} - (x + z)s + xz$

$ \Rightarrow s(x + z - y) = xz$

$ \Rightarrow {(x + z)^2} - {y^2} = 2xz$

$ \Rightarrow {x^2} + {z^2} = {y^2} \Rightarrow y = {\pi \over 2}$

$ \because $ $X + Y + Z = \pi \Rightarrow X + Z = {\pi \over 2} = Y$

$ \Rightarrow X + Z = Y$

$ \because $ $\tan {X \over 2} = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} = \sqrt {{{1 - {z \over y}} \over {1 + {z \over y}}}} $



$ = \sqrt {{{y - z} \over {y + z}}} = {{\sqrt {{y^2} - {z^2}} } \over {y + z}}$

$\tan {X \over 2} = {x \over {y + z}}$ {$ \because $ x² + z² = y²}

Let a non-right angled $\Delta $PQR.

Now, by sine rule



${P \over {\sin P}} = {q \over {\sin Q}} = {r \over {\sin R}} = 2 \times $ circumradius

$ \Rightarrow {{\sqrt 3 } \over {\sin P}} = {1 \over {\sin Q}} = {r \over {\sin R}} = 2 \times 1$

[circumradius = 1 unit]

$ \Rightarrow $ ${\sin P}$ = ${{\sqrt 3 } \over 2}$ and sinQ = ${1 \over 2}$

$ \Rightarrow P = 120^\circ $ and $ \Rightarrow Q = 30^\circ $

($ \because $ $\Delta $PQR is non-right angled triangle)

So, R = 30$^\circ $

$ \Rightarrow $ r = 1, so $\Delta $PQR is an isosceles triangle. And, since RS and PE are the median of $\Delta $PQR, so 'O' is centroid of the $\Delta $PQR.

Now,

Option (a),

From Apollonius theorem,

$2(P{E^2} + Q{E^2}) = P{Q^2} + P{R^2}$

$ \Rightarrow 2\left( {P{E^2} + {3 \over 4}} \right) = 1 + 1$

$ \Rightarrow P{E^2} = 1 - {3 \over 4} \Rightarrow P{E^2} = {1 \over 4}$

$ \Rightarrow PE = {1 \over 2}$ units

and $OE = {1 \over 3}PE = {1 \over 6}$ units

[$ \because $ O divides PE is 2 : 1]

Option (b),

Again from Apollonius theorem,

2 (PS² + RS²) = PR² + QR²

$ \Rightarrow 2\left( {{1 \over 4} + R{S^2}} \right) = 1 + 3$

$ \Rightarrow R{S^2} = 2 - {1 \over 4} \Rightarrow R{S^2} = {7 \over 4}$

$ \Rightarrow RS = {{\sqrt 7 } \over 2}$ units

Option (c),

Area of $\Delta $SOE = ${1 \over 2}(OE)\,(ST)$

= ${1 \over 2} \times {1 \over 6}[(PS)sin60^\circ ]$

= ${1 \over {12}} \times {1 \over 2} \times {{\sqrt 3 } \over 2}$

= ${{\sqrt 3 } \over {48}}$ square units

Option (d),

$ \because $ Inradius of

$\Delta PQR = {\Delta \over s} = {{{1 \over 2}pq\sin R} \over {{1 \over 2}(p + q + r)}} = {{{1 \over 2}(\sqrt 2 )(1){1 \over 2}} \over {{1 \over 2}(\sqrt 3 + 1 + 1)}}$

$ = {{\sqrt 3 } \over 2}(2 - \sqrt 3 )$ units

Hence, options (a), (b) and (d) are correct.

It is given that

$ 2 s=x+y+z $

Let us consider

$ \frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}=k $

That is, $s=4 k+x$

$ \begin{aligned} &s =3 k+y \\\\ &s =2 k+z \\\\ &3 s =9 k+(x+y+z)=9 k+2 s \\\\ &s =9 k \\\\ &x =9 k-4 k=5 k \\\\ &y =6 k, z=7 k \end{aligned} $


Area of in-circle of the triangle $X Y Z$ is

$ \begin{aligned} \pi r^2 & =\pi\left(\frac{\Delta}{s}\right)^2 \\\\ & =\frac{\pi}{s^2} \Delta^2=\frac{\pi}{s^2} s(s-x)(s-y)(s-z) \\\\ & =\frac{\pi}{81 k^2} \times 9 k \times 4 k \times 3 k \times 2 k \\\\ & =\frac{\pi}{9 k} \times 24 k^3=\frac{\pi}{9} 24 k^2 \end{aligned} $

$ \begin{aligned} & \text {Therefore, } \frac{\pi}{9} 24 k^2=\frac{8 \pi}{3} \\\\ & \Rightarrow \quad k^2=\frac{8 \pi}{3} \times \frac{9}{24 \pi}=1 \Rightarrow k=1 \end{aligned} $

The sides of the triangle is given by $x=5, y=6, z=7$. Now, the area of $\triangle X Y Z$ is

$ \sqrt{s(s-x)(s-y)(s-z)}=\sqrt{9 \times 4 \times 3 \times 2}=6 \sqrt{6} $

Hence, option $(\mathrm{A})$ is correct.

Now, $R=\frac{x y z}{4 \Delta}=\frac{5 \times 6 \times 7}{4 \times 6 \sqrt{6}}=\frac{35}{4 \sqrt{6}}$

$ \begin{aligned} & \sin \frac{X}{2} \sin \frac{Y}{2} \sin \frac{Z}{2} \\\\ & \left.\quad=\sqrt{\left[\frac{(s-y)(s-z)}{y z}\right]\left[\frac{(s-z)(s-z)}{x z}\right]\left[\frac{(s-y)(s-x)}{x y}\right]}\right. \\\\ & \quad=\frac{(s-z)(s-y)(s-z)}{x y z} \\\\ & \text {Now, } \frac{(s-z)(s-y)(s-z)}{x y z}=\frac{4 \times 3 \times 2}{5 \times 6 \times 7}=\frac{4}{25} \end{aligned} $

Hence, option (C) is correct.

Now, $\sin ^2\left(\frac{X+Y}{2}\right)=\sin ^2\left(\frac{\pi-Z}{2}\right)=\cos ^2 \frac{z}{2}=\frac{s(s-z)}{x y}$

$ =\frac{9 \times 2}{5 \times 6}=\frac{3}{5} $

Hence, option (D) is correct.

For the above circle PN and PM are its pair of tangents. So, its length are equal i.e. $|\mathrm{PN}|=|\mathrm{PM}| \text {. }$

Similarly $|\mathrm{QL}|=|\mathrm{QN}|$ and $|\mathrm{RL}|=|\mathrm{RM}|$

Let $|\mathrm{PN}|=|\mathrm{PM}|=x,|\mathrm{QL}|=|\mathrm{QN}|=y$ and $|R L|=|R M|=z$

Given $|\mathrm{PN}|,|\mathrm{QL}|$ and $|\mathrm{RM}|$ are consecutive even integer

$\therefore \quad x=2 n, y=2 n+2 \text { and } z=2 n+4$

Given, $\cos P=\frac{1}{3}$

$\begin{aligned} & \Rightarrow \quad \frac{(4 n+4)^2+(4 n+2)^2-(4 n+6)^2}{2(4 n+4)(4 n+2)}=\frac{1}{3} \\ & \left(16 n^2+16+32 n\right)+\left(16 n^2+4+16 n\right) \\ & \Rightarrow \frac{-\left(16 n^2+36+48 n\right)}{32 n^2+48 n+16}=\frac{1}{3} \\ & \Rightarrow \quad 48 n^2-48=32 n^2+48 n+16 \\ & \Rightarrow \quad 16 n^2-48 n-64=0 \\ & \Rightarrow \quad n^2-3 n-4=0 \\ & \Rightarrow \quad n^2-4 n+n-4=0 \\ & \Rightarrow \quad(n-4)(n+1)=0 \\ & \Rightarrow \quad n=4 \\ \end{aligned}$

$\begin{aligned} & \text { Now }|P Q|=4 n+2=18,|Q R|=4 n+6=22 \\ & \text { and }|R P|=4 n+4=20 \end{aligned}$

Hints :

(i) Recall cosine rule in a triangle $\triangle \mathrm{ABC}$,

$\begin{aligned} & \cos \mathrm{A}=\frac{b^2+c^2-a^2}{2 b c}, \cos \mathrm{B}=\frac{c^2+a^2-b^2}{2 c a} \\ \text{and} \quad & \cos \mathrm{C}=\frac{a^2+b^2-c^2}{2 a b}\end{aligned}$

(ii) Three consecutive even integers are $2 n, 2 n +2,2 n+4$ where $n$ is an integer.

$ \text { Area of } \triangle \mathrm{ABC} $

$ \begin{aligned} & \text { = area }(\Delta \mathrm{ABD})+\text { area }(\Delta \mathrm{ACD}) \\ & \text { Let } \mathrm{AB}=c, \mathrm{BC}=a \text { and } \mathrm{AC}=b \\ & \frac{1}{2} b c \sin \mathrm{~A}=\frac{1}{2} c \times \mathrm{AD} \sin \frac{\mathrm{~A}}{2} \\ & +\frac{1}{2} \times b \times \mathrm{AD} \sin \frac{\mathrm{~A}}{2} \\ & \Rightarrow b c \sin \mathrm{~A}=c \times \mathrm{AD} \sin \frac{\mathrm{~A}}{2}+b \mathrm{AD} \sin \frac{\mathrm{~A}}{2} \\ & \Rightarrow b c 2 \sin \frac{\mathrm{~A}}{2} \cdot \cos \frac{\mathrm{~A}}{2}=c \times \mathrm{AD} \\ & \sin \frac{\mathrm{~A}}{2}+b \mathrm{AD} \sin \frac{\mathrm{~A}}{2} \\ & \because \quad \left(\sin \mathrm{~A}=2 \sin \frac{\mathrm{~A}}{2} \cos \frac{\mathrm{~A}}{2}\right) \\ & \Rightarrow \quad 2 b c \cos \frac{\mathrm{~A}}{2}=c \times \mathrm{AD}+b \times \mathrm{AD} \end{aligned} $

$ \Rightarrow \quad \mathrm{AD}=\frac{2 b c \cos \frac{\mathrm{~A}}{2}}{b+c} $

Option (B) is correct.

In $\triangle \mathrm{ADE}$,

$ \begin{aligned} \cos \left(\frac{\mathrm{A}}{2}\right)= & \frac{\mathrm{AD}}{\mathrm{AE}} \\ \mathrm{AE}= & \frac{\mathrm{AD}}{\cos \left(\frac{\mathrm{~A}}{2}\right)}=\frac{2 b c \cos \left(\frac{\mathrm{~A}}{2}\right)}{(b+c)\left(\cos \frac{\mathrm{A}}{2}\right)} \\ & \text { (using above calculated value of } \mathrm{AD} \text { ) } \\ = & \frac{2 b c}{b+c} \\ \mathrm{AE}= & \frac{2}{\frac{1}{b}+\frac{1}{c}} \text { Harmonic mean of } b \text { and } c \end{aligned} $

Option (A) is correct.

$ \begin{aligned} &\begin{aligned} \mathrm{EF} & =\mathrm{ED}+\mathrm{DF} \\ & =2 \mathrm{DE} \end{aligned}\\ &\text { Since, } \Delta \mathrm{ADE} \text { and } \Delta \mathrm{ADF} \text { is congruent by ASA }\\ &\begin{aligned} & =2 \times \mathrm{AD} \tan \left(\frac{\mathrm{~A}}{2}\right) \\ & =2 \times \frac{2 b c}{b+c} \cos \left(\frac{\mathrm{~A}}{2}\right) \frac{\sin \left(\frac{\mathrm{A}}{2}\right)}{\cos \left(\frac{\mathrm{A}}{2}\right)} \end{aligned} \end{aligned} $

$ \begin{aligned} \angle \mathrm{DAE} & =\angle \mathrm{DAF}=\angle \frac{\mathrm{A}}{2} \\ \angle \mathrm{ADE} & =\angle \mathrm{ADF}=90^{\circ} \\ \text { and } \mathrm{AD} & =\mathrm{AD}(\text { common side }) \\ \Rightarrow \mathrm{DE} & =\mathrm{DF} \text { by } \mathrm{CPCT} \\ \mathrm{EF} & =\frac{4 b c}{b+c} \sin \left(\frac{\mathrm{~A}}{2}\right) \end{aligned} $

Option (C) is correct.

Since, AD is bisector and perpendicular on side EF

Therefore, $\triangle \mathrm{AEF}$ is an isosceles.