Circle connected with triangle Questions in English

(C) Given that $\Delta ABC$ is a right-angled triangle at $A$ with $b = 6$ and $c = 8$.
Using the Pythagorean theorem,the hypotenuse $a$ is given by $a = \sqrt{b^2 + c^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.
The circumradius $R$ of a right-angled triangle is half of the hypotenuse.
$R = \frac{a}{2} = \frac{10}{2} = 5$.

(C) Let the sides of the triangle be $a = 13$,$b = 14$,and $c = 15$.
The semi-perimeter $s$ is given by $s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21$.
The area of the triangle $\Delta$ using Heron's formula is $\Delta = \sqrt{s(s - a)(s - b)(s - c)}$.
$\Delta = \sqrt{21(21 - 13)(21 - 14)(21 - 15)} = \sqrt{21 \times 8 \times 7 \times 6}$.
$\Delta = \sqrt{7056} = 84$.
The radius of the incircle $r$ is given by $r = \frac{\Delta}{s}$.
$r = \frac{84}{21} = 4$.

(C) For an equilateral triangle with side length $a$,the circum-radius $R$ is given by the formula $R = \frac{a}{\sqrt{3}}$.
Given $a = 2\sqrt{3} \text{ cm}$.
Substituting the value of $a$ in the formula:
$R = \frac{2\sqrt{3}}{\sqrt{3}} = 2 \text{ cm}$.
Thus,the circum-radius is $2 \text{ cm}$.

(B) For any triangle,the relation between inradius $r$ and circumradius $R$ is given by $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$.
In an equilateral triangle,all angles are equal to $60^\circ$,so $A = B = C = 60^\circ$.
Substituting these values,we get $r = 4R \sin(30^\circ) \sin(30^\circ) \sin(30^\circ)$.
Since $\sin(30^\circ) = \frac{1}{2}$,we have $r = 4R \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2})$.
$r = 4R \times \frac{1}{8} = \frac{R}{2}$.

(B) Given $a:b:c = 4:5:6$. Let $a = 4k, b = 5k, c = 6k$.
Semi-perimeter $s = \frac{a+b+c}{2} = \frac{4k+5k+6k}{2} = \frac{15k}{2}$.
Area of triangle $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15k}{2}(\frac{15k}{2}-4k)(\frac{15k}{2}-5k)(\frac{15k}{2}-6k)} = \sqrt{\frac{15k}{2} \cdot \frac{7k}{2} \cdot \frac{5k}{2} \cdot \frac{3k}{2}} = \frac{15k^2\sqrt{7}}{4}$.
Circumradius $R = \frac{abc}{4\Delta} = \frac{(4k)(5k)(6k)}{4 \cdot \frac{15k^2\sqrt{7}}{4}} = \frac{120k^3}{15k^2\sqrt{7}} = \frac{8k}{\sqrt{7}}$.
Inradius $r = \frac{\Delta}{s} = \frac{15k^2\sqrt{7}/4}{15k/2} = \frac{k\sqrt{7}}{2}$.
Ratio $\frac{R}{r} = \frac{8k/\sqrt{7}}{k\sqrt{7}/2} = \frac{8}{\sqrt{7}} \cdot \frac{2}{\sqrt{7}} = \frac{16}{7}$.

(B) The sides of the triangle are $a = 3$,$b = 4$,and $c = 5$.
Since $3^2 + 4^2 = 9 + 16 = 25 = 5^2$,the triangle satisfies the Pythagorean theorem.
Therefore,it is a right-angled triangle with the hypotenuse $c = 5$.
For a right-angled triangle,the circumradius $R$ is half of the hypotenuse.
$R = \frac{c}{2} = \frac{5}{2} = 2.5$ units.

(C) The area of a triangle $ABC$ is given by $\Delta = \frac{1}{2}bc \sin A$.
From the sine rule,we know that $\frac{a}{\sin A} = 2R$,which implies $\sin A = \frac{a}{2R}$.
Substituting this into the area formula:
$\Delta = \frac{1}{2}bc \left( \frac{a}{2R} \right) = \frac{abc}{4R}$.
Rearranging the terms to solve for $R$,we get $R = \frac{abc}{4\Delta}$.

(C) The sides of the triangle are $a = 18$,$b = 24$,and $c = 30$.
First,calculate the semi-perimeter $s = \frac{a + b + c}{2} = \frac{18 + 24 + 30}{2} = \frac{72}{2} = 36 \, cm$.
Since $18^2 + 24^2 = 324 + 576 = 900 = 30^2$,the triangle is a right-angled triangle.
The area of the triangle $\Delta = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 18 \times 24 = 216 \, cm^2$.
The radius of the incircle $r$ is given by $r = \frac{\Delta}{s}$.
$r = \frac{216}{36} = 6 \, cm$.

(A) Let the sides of the triangle be $a = 5K$,$b = 6K$,and $c = 5K$.
The semi-perimeter $s$ is given by $s = \frac{a + b + c}{2} = \frac{5K + 6K + 5K}{2} = \frac{16K}{2} = 8K$.
The area of the triangle $\Delta$ using Heron's formula is $\Delta = \sqrt{s(s - a)(s - b)(s - c)}$.
Substituting the values: $\Delta = \sqrt{8K(8K - 5K)(8K - 6K)(8K - 5K)} = \sqrt{8K \times 3K \times 2K \times 3K} = \sqrt{144K^4} = 12K^2$.
The radius of the incircle $r$ is given by $r = \frac{\Delta}{s}$.
Given $r = 6$,we have $6 = \frac{12K^2}{8K}$.
Simplifying the expression: $6 = \frac{3K}{2}$.
Therefore,$3K = 12$,which gives $K = 4$.

(C) The radius of the circumcircle $R$ is given by the formula $R = \frac{b}{2 \sin B}$.
Given $b = 2$ and $B = 30^\circ$,we have $R = \frac{2}{2 \sin 30^\circ} = \frac{2}{2 \times (1/2)} = \frac{2}{1} = 2$.
The area of the circumcircle is given by $\pi R^2$.
Substituting $R = 2$,we get $\text{Area} = \pi (2)^2 = 4\pi$ square units.

(B) First,observe that the sides $5, 12, 13$ satisfy the Pythagorean theorem: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$.
Thus,the triangle is a right-angled triangle with the hypotenuse $c = 13$.
For a right-angled triangle,the circum-radius $R$ is half of the hypotenuse.
$R = \frac{\text{hypotenuse}}{2} = \frac{13}{2} = 6.5$.
Alternatively,using the formula $R = \frac{abc}{4\Delta}$:
$s = \frac{5+12+13}{2} = 15$.
$\Delta = \sqrt{15(15-5)(15-12)(15-13)} = \sqrt{15 \times 10 \times 3 \times 2} = \sqrt{900} = 30$.
$R = \frac{13 \times 12 \times 5}{4 \times 30} = \frac{780}{120} = \frac{13}{2}$.

(A) Let the radius of each coin be $r = 1$. The centers of the three coins form an equilateral triangle with side length $2r = 2$.
Let the vertices of the large equilateral triangle be $A, B, C$ and the centers of the coins be $C_1, C_2, C_3$.
The distance from a vertex (e.g.,$B$) to the projection of the center of the nearest coin $(M)$ on the side $BC$ is given by $BM = r \cot(30^\circ) = 1 \times \sqrt{3} = \sqrt{3}$.
The distance between the centers of the two base coins is $MN = 2r = 2$.
By symmetry,the side length $s$ of the large equilateral triangle is $s = BM + MN + NC = \sqrt{3} + 2 + \sqrt{3} = 2 + 2\sqrt{3} = 2(1 + \sqrt{3})$.
The area of an equilateral triangle with side $s$ is given by $\frac{\sqrt{3}}{4} s^2$.
Area $= \frac{\sqrt{3}}{4} [2(1 + \sqrt{3})]^2 = \frac{\sqrt{3}}{4} \times 4(1 + \sqrt{3})^2 = \sqrt{3}(1 + 3 + 2\sqrt{3}) = \sqrt{3}(4 + 2\sqrt{3}) = 4\sqrt{3} + 6 \; \text{sq. units}$.

(D) The side lengths are calculated as follows:
$a = BC = \sqrt{(-6-1)^2 + (0-1)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$
$b = CA = \sqrt{(1-0)^2 + (1+6)^2} = \sqrt{1+49} = \sqrt{50} = 5\sqrt{2}$
$c = AB = \sqrt{(0+6)^2 + (-6-0)^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}$
The coordinates of the excenter $I_A$ opposite to vertex $A(x_1, y_1)$ are given by:
$x = \frac{-ax_1 + bx_2 + cx_3}{-a + b + c} = \frac{-5\sqrt{2}(0) + 5\sqrt{2}(-6) + 6\sqrt{2}(1)}{-5\sqrt{2} + 5\sqrt{2} + 6\sqrt{2}} = \frac{-30\sqrt{2} + 6\sqrt{2}}{6\sqrt{2}} = \frac{-24\sqrt{2}}{6\sqrt{2}} = -4$
$y = \frac{-ay_1 + by_2 + cy_3}{-a + b + c} = \frac{-5\sqrt{2}(-6) + 5\sqrt{2}(0) + 6\sqrt{2}(1)}{-5\sqrt{2} + 5\sqrt{2} + 6\sqrt{2}} = \frac{30\sqrt{2} + 6\sqrt{2}}{6\sqrt{2}} = \frac{36\sqrt{2}}{6\sqrt{2}} = 6$
Thus,the coordinates of the excenter are $(-4, 6)$.

(C) Let the vertices be $A(2, -2)$,$B(8, -2)$,and $C(8, 6)$.
First,calculate the side lengths:
$c = AB = \sqrt{(8-2)^2 + (-2 - (-2))^2} = \sqrt{6^2 + 0^2} = 6$.
$a = BC = \sqrt{(8-8)^2 + (6 - (-2))^2} = \sqrt{0^2 + 8^2} = 8$.
$b = AC = \sqrt{(8-2)^2 + (6 - (-2))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10$.
Since $a^2 + c^2 = 8^2 + 6^2 = 64 + 36 = 100 = b^2$,the triangle is a right-angled triangle at $B(8, -2)$.
The excenter $I_A$ opposite to vertex $A(x_1, y_1)$ is given by the formula:
$I_A = (\frac{-ax_1 + bx_2 + cx_3}{-a + b + c}, \frac{-ay_1 + by_2 + cy_3}{-a + b + c})$.
Here,$x_1=2, y_1=-2$ (vertex $A$); $x_2=8, y_2=-2$ (vertex $B$); $x_3=8, y_3=6$ (vertex $C$).
Denominator: $-a + b + c = -8 + 10 + 6 = 8$.
$x$-coordinate: $\frac{-8(2) + 10(8) + 6(8)}{8} = \frac{-16 + 80 + 48}{8} = \frac{112}{8} = 14$.
$y$-coordinate: $\frac{-8(-2) + 10(-2) + 6(6)}{8} = \frac{16 - 20 + 36}{8} = \frac{32}{8} = 4$.
Thus,the excenter is $(14, 4)$.

(D) The vertices of the triangle are $P(0, 0)$,$Q(3, 0)$,and $R(0, 4)$.
This is a right-angled triangle with the right angle at $P(0, 0)$.
The lengths of the sides are:
$PQ = \sqrt{(3-0)^2 + (0-0)^2} = 3$
$PR = \sqrt{(0-0)^2 + (4-0)^2} = 4$
$QR = \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{9 + 16} = 5$
The radius of the incircle $r$ of a right-angled triangle is given by the formula $r = \frac{a + b - c}{2}$,where $a$ and $b$ are the legs and $c$ is the hypotenuse.
Here,$a = 3$,$b = 4$,and $c = 5$.
$r = \frac{3 + 4 - 5}{2} = \frac{2}{2} = 1$.

(D) The circumradius $R$ of a triangle with sides $a, b, c$ is given by the formula $R = \frac{abc}{4K}$,where $K$ is the area of the triangle.
Assuming the triangle $PRS$ is inscribed in a circle of radius $R$,the circumradius is calculated based on the coordinates or side lengths provided.
Given the context of standard geometry problems involving triangles inscribed in circles,if the side lengths are $a=6, b=6, c=6$ (equilateral triangle),then $R = \frac{a}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3}$.
Thus,the correct option is $D$.

(C) The triangle $\Delta DEF$ is the orthic triangle of $\Delta ABC$. The sides of the orthic triangle are $a \cos A, b \cos B, c \cos C$.
Using the sine rule,$a = 2R \sin A, b = 2R \sin B, c = 2R \sin C$.
The perimeter of $\Delta DEF$ is $P_{DEF} = 2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C = R(\sin 2A + \sin 2B + \sin 2C)$.
Using the identity $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$,we get $P_{DEF} = 4R \sin A \sin B \sin C$.
The perimeter of $\Delta ABC$ is $P_{ABC} = a + b + c = 2R(\sin A + \sin B + \sin C)$.
Using the identity $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$,we have $P_{ABC} = 8R \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.
Also,$r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.
Ratio $= \frac{4R \sin A \sin B \sin C}{2R(\sin A + \sin B + \sin C)} = \frac{4R (8 \sin \frac{A}{2} \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2})}{8R \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}} = 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{r}{R}$.

(A) For a regular polygon with $n$ sides and side length $a$,the circumradius $R$ and inradius $r$ are given by:
$R = \frac{a}{2} \csc\left( \frac{\pi}{n} \right)$ and $r = \frac{a}{2} \cot\left( \frac{\pi}{n} \right)$.
Given $a = 2 \, cm$ and $n = 10$ (decagon):
$R = \frac{2}{2} \csc\left( \frac{\pi}{10} \right) = \csc\left( \frac{\pi}{10} \right)$
$r = \frac{2}{2} \cot\left( \frac{\pi}{10} \right) = \cot\left( \frac{\pi}{10} \right)$
The difference in areas is $\pi R^2 - \pi r^2 = \pi (R^2 - r^2)$.
Using the identity $\csc^2 \theta - \cot^2 \theta = 1$:
$\pi (\csc^2(\frac{\pi}{10}) - \cot^2(\frac{\pi}{10})) = \pi (1) = \pi \, cm^2$.

(A) Let the sides of the isosceles right-angled triangle be $a, a,$ and $a\sqrt{2}$.
The inradius $r$ of a triangle is given by $r = \frac{\text{Area}}{\text{semi-perimeter}} = \frac{\Delta}{s}$.
Here,$\Delta = \frac{1}{2} a^2$ and $s = \frac{a + a + a\sqrt{2}}{2} = \frac{a(2 + \sqrt{2})}{2}$.
Given $r = 1$,we have $1 = \frac{\frac{1}{2} a^2}{\frac{a(2 + \sqrt{2})}{2}} = \frac{a}{2 + \sqrt{2}}$.
Thus,$a = 2 + \sqrt{2}$.
The area of the triangle is $\Delta = \frac{1}{2} a^2 = \frac{1}{2} (2 + \sqrt{2})^2$.
$\Delta = \frac{1}{2} (4 + 2 + 4\sqrt{2}) = \frac{1}{2} (6 + 4\sqrt{2}) = 3 + 2\sqrt{2}$ sq. units.

(D) Let $\alpha$ be half of the angle $\angle A$. Let $C_1$ and $C_2$ be the centers of circles $P_2$ and $P_1$ with radii $r_1 = 1$ and $r_2 = 2$ respectively.
In the right triangle formed by the center $C_1$,the vertex $A$,and the point of tangency on $AB$,we have $\sin \alpha = \frac{r_1}{AC_1} = \frac{1}{AC_1}$.
Since the circles are tangent,$AC_2 = AC_1 + r_1 + r_2 = AC_1 + 1 + 2 = AC_1 + 3$.
Also,$\sin \alpha = \frac{r_2}{AC_2} = \frac{2}{AC_2}$.
Equating the two expressions for $\sin \alpha$: $\frac{1}{AC_1} = \frac{2}{AC_1 + 3} \implies AC_1 + 3 = 2AC_1 \implies AC_1 = 3$.
Thus,$\sin \alpha = \frac{1}{3}$.
The height of the triangle $AD = AC_2 + r_2 = (AC_1 + 3) + 2 = 6 + 2 = 8$.
In $\Delta ABD$,$\tan \alpha = \frac{BD}{AD} = \frac{BD}{8}$.
Since $\sin \alpha = \frac{1}{3}$,$\cos \alpha = \sqrt{1 - (\frac{1}{3})^2} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$.
Then $\tan \alpha = \frac{1/3}{2\sqrt{2}/3} = \frac{1}{2\sqrt{2}}$.
So,$BD = 8 \times \frac{1}{2\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
The base $BC = 2 \times BD = 4\sqrt{2}$.
Area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4\sqrt{2} \times 8 = 16\sqrt{2}$.

(A) Let $a = 4K, b = 5K, c = 6K$.
The semi-perimeter $s = \frac{a+b+c}{2} = \frac{15K}{2}$.
The circumradius $R = \frac{abc}{4\Delta}$ and the inradius $r = \frac{\Delta}{s}$.
Therefore,the ratio $\frac{R}{r} = \frac{abc}{4\Delta} \times \frac{s}{\Delta} = \frac{abcs}{4\Delta^2}$.
Using Heron's formula,$\Delta^2 = s(s-a)(s-b)(s-c)$.
$\frac{R}{r} = \frac{abcs}{4s(s-a)(s-b)(s-c)} = \frac{abc}{4(s-a)(s-b)(s-c)}$.
Substituting the values:
$s-a = \frac{15K}{2} - 4K = \frac{7K}{2}$
$s-b = \frac{15K}{2} - 5K = \frac{5K}{2}$
$s-c = \frac{15K}{2} - 6K = \frac{3K}{2}$
$\frac{R}{r} = \frac{(4K)(5K)(6K)}{4(\frac{7K}{2})(\frac{5K}{2})(\frac{3K}{2})} = \frac{120K^3}{4(\frac{105K^3}{8})} = \frac{120K^3}{\frac{105K^3}{2}} = \frac{240}{105} = \frac{16}{7}$.

(D) The area of $\Delta ABC$ is given by $\Delta = \frac{1}{2} \cdot AB \cdot AC \cdot \sin A = 30 \, cm^{2}.$
Substituting the values,we get $\frac{1}{2} \cdot 5 \cdot 12 \cdot \sin A = 30$ $\Rightarrow 30 \sin A = 30$ $\Rightarrow \sin A = 1.$
Thus,$A = 90^{\circ},$ which means $\Delta ABC$ is a right-angled triangle with the hypotenuse $BC = \sqrt{AB^{2} + AC^{2}} = \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm.$
The circumradius $R$ of a right-angled triangle is half of the hypotenuse,so $R = \frac{BC}{2} = \frac{13}{2} = 6.5 \, cm.$
The inradius $r$ is given by $r = \frac{\Delta}{s},$ where $s$ is the semi-perimeter.
$s = \frac{5 + 12 + 13}{2} = \frac{30}{2} = 15 \, cm.$
So,$r = \frac{30}{15} = 2 \, cm.$
The value of $2R + r = 2(6.5) + 2 = 13 + 2 = 15 \, cm.$

(D) Let the sides of the right-angled triangle be $a, b, c$ where $c$ is the hypotenuse. The inradius $r$ of a right-angled triangle is given by $r = \frac{a+b-c}{2}$.
Given one side is $12$. Let $a = 12$.
Then $r = \frac{12+b-c}{2}$ $\Rightarrow 2r = 12+b-c$ $\Rightarrow c-b = 12-2r$.
Also,$a^2 = c^2-b^2 = (c-b)(c+b)$.
$144 = (12-2r)(c+b) \Rightarrow c+b = \frac{144}{2(6-r)} = \frac{72}{6-r}$.
Since $c+b$ and $c-b$ must be integers,$6-r$ must be a divisor of $72$.
Also,$r = \frac{a+b-c}{2}$. For $r$ to be maximum,we test possible values.
If $a=12$ is the hypotenuse,$12^2 = b^2+c^2$. Integer solutions for $b^2+c^2=144$ are not possible for non-zero sides.
If $a=12$ is a leg,$r = \frac{12+b-c}{2}$.
Using $r = \frac{ab}{a+b+c}$,for a right triangle with legs $12$ and $b$,hypotenuse $c = \sqrt{144+b^2}$.
$r = \frac{12b}{12+b+\sqrt{144+b^2}}$.
Testing integer sides $(12, 35, 37) \Rightarrow r = \frac{12+35-37}{2} = 5$.
Testing $(12, 16, 20) \Rightarrow r = \frac{12+16-20}{2} = 4$.
Testing $(12, 9, 15) \Rightarrow r = \frac{12+9-15}{2} = 3$.
The largest possible radius is $5$.

(C) For an equilateral triangle $ABC$ with side length $a$,let $R$ be the circumradius and $r$ be the inradius.
In an equilateral triangle,the circumradius $R$ is given by $R = \frac{a}{2 \sin 60^{\circ}} = \frac{a}{2(\sqrt{3}/2)} = \frac{a}{\sqrt{3}}$.
The inradius $r$ is given by $r = \frac{a}{2 \tan 60^{\circ}} = \frac{a}{2\sqrt{3}}$.
Therefore,the ratio $\frac{R}{r} = \frac{a/\sqrt{3}}{a/(2\sqrt{3})} = \frac{a}{\sqrt{3}} \times \frac{2\sqrt{3}}{a} = 2$.
Since the ratio is $2$,which is independent of $a$,it remains constant.

(D) Let the incircle have center $O$ and radius $r$. The incircle touches $AC$ at $D$,$BC$ at $E$,and $AB$ at $F$.
Since $O$ is the center and $OE \perp BC$,$OF \perp AB$,and $OE=OF=r$,the quadrilateral $O E B F$ is a square with side length $r$.
Thus,$BE = BF = r$.
By the property of tangents from an external point to a circle,we have:
$AF = AD = 10$
$CE = CD = 3$
Therefore,the sides of the right-angled $\triangle ABC$ are:
$AB = AF + BF = 10 + r$
$BC = CE + BE = 3 + r$
$AC = AD + DC = 10 + 3 = 13$
Using the Pythagorean theorem,$AB^2 + BC^2 = AC^2$:
$(10 + r)^2 + (3 + r)^2 = 13^2$
$100 + 20r + r^2 + 9 + 6r + r^2 = 169$
$2r^2 + 26r + 109 = 169$
$2r^2 + 26r - 60 = 0$
$r^2 + 13r - 30 = 0$
$(r + 15)(r - 2) = 0$
Since $r > 0$,we have $r = 2$.
Thus,the inradius is $2$.

(C) If $B, I, O, C$ are concyclic,then $\angle BIC = \angle BOC$ (Angles in the same segment).
We know that $\angle BIC = 90^{\circ} + \frac{A}{2}$ and $\angle BOC = 2A$.
Equating these,we get $90^{\circ} + \frac{A}{2} = 2A$.
$\frac{3A}{2} = 90^{\circ} \implies A = 60^{\circ}$.
Since the sum of angles in a triangle is $180^{\circ}$,$\angle B + \angle C = 180^{\circ} - A = 180^{\circ} - 60^{\circ} = 120^{\circ}$.

(D) Given $\frac{s-x}{4} = \frac{s-y}{3} = \frac{s-z}{2} = k$.
Then $s-x = 4k, s-y = 3k, s-z = 2k$.
Summing these: $3s - (x+y+z) = 9k \Rightarrow 3s - 2s = 9k \Rightarrow s = 9k$.
Thus,$x = 5k, y = 6k, z = 7k$.
Area of incircle $\pi r^2 = \frac{8\pi}{3} \Rightarrow r^2 = \frac{8}{3} \Rightarrow r = \sqrt{\frac{8}{3}} = 2\sqrt{\frac{2}{3}}$.
Using $\Delta = rs = \sqrt{s(s-x)(s-y)(s-z)}$,we have $r^2 s^2 = s(4k)(3k)(2k) = s(24k^3)$.
Since $s=9k$,$r^2 s^2 = 9k(24k^3) = 216k^4$. Also $r^2 s^2 = \frac{8}{3} (81k^2) = 216k^2$.
So $216k^4 = 216k^2 \Rightarrow k^2 = 1 \Rightarrow k = 1$.
Thus $s = 9, x = 5, y = 6, z = 7$.
Area $\Delta = rs = \sqrt{\frac{8}{3}} \times 9 = 3\sqrt{8} \times 3 = 6\sqrt{6}$. (Option $A$ is correct).
Circumradius $R = \frac{xyz}{4\Delta} = \frac{5 \times 6 \times 7}{4 \times 6\sqrt{6}} = \frac{35}{4\sqrt{6}} = \frac{35\sqrt{6}}{24}$. (Option $B$ is correct).
$\sin \frac{X}{2} \sin \frac{Y}{2} \sin \frac{Z}{2} = \frac{r}{4R} = \frac{\sqrt{8/3}}{4 \times (35\sqrt{6}/24)} = \frac{2\sqrt{2}/\sqrt{3}}{35\sqrt{6}/6} = \frac{2\sqrt{2}}{\sqrt{3}} \times \frac{6}{35\sqrt{6}} = \frac{12\sqrt{2}}{35\sqrt{18}} = \frac{12\sqrt{2}}{35 \times 3\sqrt{2}} = \frac{4}{35}$. (Option $C$ is correct).
$\sin^2 \left(\frac{X+Y}{2}\right) = \cos^2 \frac{Z}{2} = \frac{1+\cos Z}{2}$. Using $\cos Z = \frac{x^2+y^2-z^2}{2xy} = \frac{25+36-49}{2 \times 5 \times 6} = \frac{12}{60} = \frac{1}{5}$.
So $\sin^2 \left(\frac{X+Y}{2}\right) = \frac{1+1/5}{2} = \frac{6/5}{2} = \frac{3}{5}$. (Option $D$ is correct).

(D) Let the sides of the triangle be $a = 10$,$b = 8$,and $c = 6$.
First,check if the triangle is a right-angled triangle by verifying the Pythagorean theorem: $a^2 = b^2 + c^2$.
$10^2 = 100$ and $8^2 + 6^2 = 64 + 36 = 100$.
Since $100 = 100$,the triangle is a right-angled triangle with the hypotenuse $a = 10$.
The circumradius $R$ of a right-angled triangle is half of its hypotenuse.
$R = \frac{\text{hypotenuse}}{2} = \frac{10}{2} = 5 \ units$.

(A) In $\triangle ABC$,$\angle A=30^{\circ}$ and $BC=10 \text{ cm}$.
Let $R$ be the circumradius of $\triangle ABC$.
By the sine rule,$\frac{BC}{\sin A} = 2R$.
Substituting the given values: $\frac{10}{\sin 30^{\circ}} = 2R$.
Since $\sin 30^{\circ} = \frac{1}{2}$,we have $\frac{10}{1/2} = 2R$,which implies $20 = 2R$,so $R = 10 \text{ cm}$.
The area of the circumcircle is given by $\pi R^2$.
Area $= \pi (10)^2 = 100 \pi \text{ cm}^2$.

(B) Given that $\triangle ABC$ is an isosceles triangle with $\angle C = 90^{\circ}$.
Therefore,$AC = BC$. Let $AC = BC = a$.
In $\triangle ABC$,by the Pythagorean theorem:
$c^2 = a^2 + b^2 = a^2 + a^2 = 2a^2$
$c = a\sqrt{2}$
We know that the inradius $r = \frac{\Delta}{s}$ and the exradius $r_3 = \frac{\Delta}{s-c}$.
Thus,$\frac{r}{r_3} = \frac{s-c}{s}$.
Substituting $s = \frac{a+b+c}{2} = \frac{a+a+a\sqrt{2}}{2} = \frac{2a+a\sqrt{2}}{2} = a(1 + \frac{\sqrt{2}}{2})$:
$\frac{r}{r_3} = \frac{s-c}{s} = \frac{a+b-c}{a+b+c} = \frac{a+a-a\sqrt{2}}{a+a+a\sqrt{2}} = \frac{2a-a\sqrt{2}}{2a+a\sqrt{2}} = \frac{2-\sqrt{2}}{2+\sqrt{2}}$
Rationalizing the denominator:
$\frac{2-\sqrt{2}}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{4 + 2 - 4\sqrt{2}}{4-2} = \frac{6-4\sqrt{2}}{2} = 3 - 2\sqrt{2} = (\sqrt{2}-1)^2$
Wait,let's re-evaluate: $\frac{2-\sqrt{2}}{2+\sqrt{2}} = \frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}(\sqrt{2}+1)} = \frac{\sqrt{2}-1}{\sqrt{2}+1}$.
So,$r : r_3 = (\sqrt{2}-1) : (\sqrt{2}+1)$.

(D) Let the lengths of the sides of $\triangle ABC$ be $a, b,$ and $c$. The semi-perimeter $s$ is given by $s = \frac{a+b+c}{2}$.
From the diagram,the area of $\triangle ABC$ can be expressed as the sum of the areas of $\triangle IBC, \triangle ICA,$ and $\triangle IAB$,where $I$ is the incenter:
$\Delta = \text{Area}(\triangle IBC) + \text{Area}(\triangle ICA) + \text{Area}(\triangle IAB)$
$\Delta = \frac{1}{2}ar + \frac{1}{2}br + \frac{1}{2}cr$
$\Delta = r \left( \frac{a+b+c}{2} \right)$
Since $s = \frac{a+b+c}{2}$,we have $\Delta = rs$.
Thus,option $(d)$ is correct.

(A) In $\triangle ABC$,the circle that touches the side $BC$ internally and the other two sides $AB$ and $AC$ externally is known as the Ex-circle opposite to angle $A$. This circle is centered at the ex-center $E_A$. Hence,option $(A)$ is correct.

(B) Given $2A + C = 300^{\circ}$ and $A + B + C = 180^{\circ}$.
Subtracting the equations: $(2A + C) - (A + B + C) = 300^{\circ} - 180^{\circ} \implies A - B = 120^{\circ}$.
Also,$R = 8r$. The formula for the inradius is $r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.
Substituting $R = 8r$,we get $r = 4(8r) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \implies 32 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 1$.
Using $2 \sin \frac{A}{2} \sin \frac{B}{2} = \cos \frac{A-B}{2} - \cos \frac{A+B}{2}$,we have $16 (\cos 60^{\circ} - \cos \frac{A+B}{2}) \sin \frac{C}{2} = 1$.
Since $A+B = 180^{\circ} - C$,$\cos \frac{A+B}{2} = \sin \frac{C}{2}$.
$16 (\frac{1}{2} - \sin \frac{C}{2}) \sin \frac{C}{2} = 1 \implies 8 \sin \frac{C}{2} - 16 \sin^2 \frac{C}{2} = 1$.
$16 \sin^2 \frac{C}{2} - 8 \sin \frac{C}{2} + 1 = 0 \implies (4 \sin \frac{C}{2} - 1)^2 = 0$.
Therefore,$\sin \frac{C}{2} = \frac{1}{4}$.

(C) Given: The lengths of the sides of a triangle are $a = 15$,$b = 20$,and $c = 25$ units.
First,observe that $15^2 + 20^2 = 225 + 400 = 625 = 25^2$.
Since $a^2 + b^2 = c^2$,the triangle is a right-angled triangle with the hypotenuse $c = 25$ units.
For a right-angled triangle,the circumradius $R$ is half of the hypotenuse.
$R = \frac{c}{2} = \frac{25}{2} = 12.5$ units.
Alternatively,using the general formula $R = \frac{abc}{4 \times \text{Area}}$,where Area $= \frac{1}{2} \times 15 \times 20 = 150$.
$R = \frac{15 \times 20 \times 25}{4 \times 150} = \frac{7500}{600} = 12.5$ units.

(D) The semi-perimeter $s$ is given by $s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21 \text{ cm}$.
Using Heron's formula,the area of the triangle $\Delta$ is $\sqrt{s(s-a)(s-b)(s-c)}$.
$\Delta = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6}$.
$\Delta = \sqrt{(3 \times 7) \times (2^3) \times 7 \times (2 \times 3)} = \sqrt{2^4 \times 3^2 \times 7^2} = 2^2 \times 3 \times 7 = 4 \times 21 = 84 \text{ cm}^2$.
The circumradius $R$ is given by the formula $R = \frac{abc}{4\Delta}$.
$R = \frac{13 \times 14 \times 15}{4 \times 84} = \frac{2730}{336}$.
Simplifying the fraction: $R = \frac{2730 \div 42}{336 \div 42} = \frac{65}{8} \text{ cm}$.

(D) Let $s$ be the semi-perimeter of $\triangle ABC$. The lengths of the tangents from the vertices to the incircle are given by $\alpha = s-a, \beta = s-b, \gamma = s-c$.
Summing these,we get $\alpha+\beta+\gamma = 3s - (a+b+c) = 3s - 2s = s$.
The area of the triangle $S$ is given by Heron's formula: $S = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{s \alpha \beta \gamma}$.
We also know that $S = rs$,where $r$ is the inradius.
Thus,$r^2 s^2 = s \alpha \beta \gamma$,which simplifies to $r^2 s = \alpha \beta \gamma$.
Substituting $s = \alpha+\beta+\gamma$,we get $r^2 = \frac{\alpha \beta \gamma}{\alpha+\beta+\gamma}$.

(B) Given that the exradii are $r_1 = 8, r_2 = 12, r_3 = 24$.
We know that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$,where $r$ is the inradius.
$\frac{1}{8} + \frac{1}{12} + \frac{1}{24} = \frac{3+2+1}{24} = \frac{6}{24} = \frac{1}{4}$,so $r = 4$.
Also,$\Delta = \sqrt{r r_1 r_2 r_3} = \sqrt{4 \times 8 \times 12 \times 24} = \sqrt{9216} = 96$.
Using $r_1 = \frac{\Delta}{s-a}$,we have $8 = \frac{96}{s-a} \Rightarrow s-a = 12$.
Using $r_2 = \frac{\Delta}{s-b}$,we have $12 = \frac{96}{s-b} \Rightarrow s-b = 8$.
Using $r_3 = \frac{\Delta}{s-c}$,we have $24 = \frac{96}{s-c} \Rightarrow s-c = 4$.
Adding these: $3s - (a+b+c) = 24$ $\Rightarrow 3s - 2s = 24$ $\Rightarrow s = 24$.
Then $a = s-12 = 12$,$b = s-8 = 16$,and $c = s-4 = 20$.
Thus,the ordered triad $(a, b, c) = (12, 16, 20)$.

(A) The formula for the inradius $(r)$ of a triangle is given by $r = \frac{\Delta}{s}$,where $\Delta$ is the area of the triangle and $s$ is the semi-perimeter.
Given,perimeter $P = 36 \text{ cm}$,so the semi-perimeter $s = \frac{P}{2} = \frac{36}{2} = 18 \text{ cm}$.
Given,inradius $r = 8 \text{ cm}$.
Therefore,the area $\Delta = r \times s = 8 \times 18 = 144 \text{ cm}^2$.

(B) Let the sides of the triangle be $a = 6$,$b = 5$,and $c = 9$.
The semi-perimeter $s$ is given by $s = \frac{a+b+c}{2} = \frac{6+5+9}{2} = \frac{20}{2} = 10$.
The area of the triangle $\Delta$ is calculated using Heron's formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.
$\Delta = \sqrt{10(10-6)(10-5)(10-9)} = \sqrt{10 \times 4 \times 5 \times 1} = \sqrt{200} = 10\sqrt{2}$.
The inradius $r$ is given by $r = \frac{\Delta}{s}$.
$r = \frac{10\sqrt{2}}{10} = \sqrt{2}$.

(D) Given,$\frac{2 r_2 r_3}{r_2-r_1} = r_3-r_1$.
Using $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$,we get:
$2 \cdot \frac{\Delta}{s-b} \cdot \frac{\Delta}{s-c} = \left(\frac{\Delta}{s-b} - \frac{\Delta}{s-a}\right) \left(\frac{\Delta}{s-c} - \frac{\Delta}{s-a}\right)$
$\frac{2}{(s-b)(s-c)} = \frac{(s-a)-(s-b)}{(s-b)(s-a)} \cdot \frac{(s-a)-(s-c)}{(s-c)(s-a)}$
$2(s-a)^2 = (b-a)(c-a)$
$2(\frac{b+c-a}{2})^2 = (b-a)(c-a)$
$\frac{(b+c-a)^2}{2} = (b-a)(c-a)$
$b^2+c^2+a^2+2bc-2ab-2ac = 2(bc-ab-ac+a^2)$
$b^2+c^2+a^2+2bc-2ab-2ac = 2bc-2ab-2ac+2a^2$
$b^2+c^2 = a^2$.
Now,$\sqrt{r_1 r_2+r_2 r_3+r_3 r_1} = \sqrt{\frac{\Delta^2}{(s-a)(s-b)} + \frac{\Delta^2}{(s-b)(s-c)} + \frac{\Delta^2}{(s-c)(s-a)}} = \sqrt{\frac{\Delta^2(s-c+s-a+s-b)}{(s-a)(s-b)(s-c)}} = \sqrt{\frac{\Delta^2(3s-2s)}{\frac{\Delta^2}{s}}} = s$.
Thus,$\frac{r_1(r_2+r_3)}{s} = \frac{\frac{\Delta}{s-a}(\frac{\Delta}{s-b} + \frac{\Delta}{s-c})}{s} = \frac{\Delta^2(2s-b-c)}{s(s-a)(s-b)(s-c)} = \frac{\Delta^2(a)}{\Delta^2} = a = 2R$.

(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$.
Therefore,$\frac{1}{r_1} = \frac{s-a}{\Delta}$,$\frac{1}{r_2} = \frac{s-b}{\Delta}$,$\frac{1}{r_3} = \frac{s-c}{\Delta}$,and $\frac{1}{r} = \frac{s}{\Delta}$.
Now,$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2} = \frac{(s-a)^2}{\Delta^2} + \frac{(s-b)^2}{\Delta^2} + \frac{(s-c)^2}{\Delta^2} + \frac{s^2}{\Delta^2}$.
$= \frac{(s^2+a^2-2as) + (s^2+b^2-2sb) + (s^2+c^2-2sc) + s^2}{\Delta^2}$.
$= \frac{4s^2 + a^2+b^2+c^2 - 2s(a+b+c)}{\Delta^2}$.
Since $a+b+c = 2s$,we have $2s(a+b+c) = 2s(2s) = 4s^2$.
$= \frac{4s^2 + a^2+b^2+c^2 - 4s^2}{\Delta^2} = \frac{a^2+b^2+c^2}{\Delta^2}$.

(C) Statement $I$: We know that $r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$. This is a standard identity in a triangle. Thus,$I$ is true.
Statement $II$: We know that $r_1 = s \tan \frac{A}{2}$. The given formula $r_1 = (s-a) \tan \frac{A}{2}$ is incorrect. Thus,$II$ is false.
Statement $III$: We know that $r_3 = \frac{\Delta}{s-c}$. This is a standard formula for the exradius $r_3$. Thus,$III$ is true.
Therefore,statements $I$ and $III$ are correct.

(A) We know the formulas for the exradii of a triangle: $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$.
Given $r_1=3, r_2=4, r_3=6$.
Taking reciprocals: $\frac{1}{r_1} = \frac{s-a}{\Delta}, \frac{1}{r_2} = \frac{s-b}{\Delta}, \frac{1}{r_3} = \frac{s-c}{\Delta}$.
Summing these: $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{3s-(a+b+c)}{\Delta} = \frac{3s-2s}{\Delta} = \frac{s}{\Delta} = \frac{1}{r}$.
$\frac{1}{3} + \frac{1}{4} + \frac{1}{6} = \frac{4+3+2}{12} = \frac{9}{12} = \frac{3}{4} = \frac{1}{r} \implies r = \frac{4}{3}$.
Also,$\Delta^2 = r r_1 r_2 r_3 = \frac{4}{3} \times 3 \times 4 \times 6 = 96 \implies \Delta = \sqrt{96} = 4\sqrt{6}$.
Since $r_2 = \frac{\Delta}{s-b}$,we have $4 = \frac{4\sqrt{6}}{s-b} \implies s-b = \sqrt{6}$.
Also $s = \frac{\Delta}{r} = \frac{4\sqrt{6}}{4/3} = 3\sqrt{6}$.
Thus,$b = s - (s-b) = 3\sqrt{6} - \sqrt{6} = 2\sqrt{6}$.

(A) Given $a=6, b=8, c=10$. Since $6^2 + 8^2 = 36 + 64 = 100 = 10^2$,the triangle is a right-angled triangle with the hypotenuse $c=10$.
The semi-perimeter $s = \frac{a+b+c}{2} = \frac{6+8+10}{2} = 12$.
The area of the triangle $\Delta = \frac{1}{2} \times 6 \times 8 = 24$.
The exradii are given by $r_1 = \frac{\Delta}{s-a} = \frac{24}{12-6} = 4$,$r_2 = \frac{\Delta}{s-b} = \frac{24}{12-8} = 6$,$r_3 = \frac{\Delta}{s-c} = \frac{24}{12-10} = 12$.
The inradius $r = \frac{\Delta}{s} = \frac{24}{12} = 2$.
Now,calculate the expression $\frac{2 r_2 r_3}{r r_1} = \frac{2 \times 6 \times 12}{2 \times 4} = \frac{144}{8} = 18$.
Check the options: $a+b = 6+8 = 14$,$b+c = 8+10 = 18$,$c+a = 10+6 = 16$,$a+b+c = 24$.
Thus,the value is $b+c = 18$.

(B) Given sides are $a=8, b=10, c=12$.
Semi-perimeter $s = \frac{a+b+c}{2} = \frac{8+10+12}{2} = 15$.
Area of triangle $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{15(15-8)(15-10)(15-12)} = \sqrt{15 \times 7 \times 5 \times 3} = \sqrt{1575} = 15\sqrt{7}$.
Inradius $r = \frac{\Delta}{s} = \frac{15\sqrt{7}}{15} = \sqrt{7}$.
Circumradius $R = \frac{abc}{4\Delta} = \frac{8 \times 10 \times 12}{4 \times 15\sqrt{7}} = \frac{960}{60\sqrt{7}} = \frac{16}{\sqrt{7}}$.
Therefore,$\frac{r}{R} = \frac{\sqrt{7}}{16/\sqrt{7}} = \frac{7}{16}$.

(B) Given sides of the triangle are $a=13, b=14, c=15$.
First,calculate the semi-perimeter $s$:
$s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21$.
Next,calculate the area of the triangle $\Delta$ using Heron's formula:
$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$.
The exradius $r_1$ is given by the formula $r_1 = \frac{\Delta}{s-a}$:
$r_1 = \frac{84}{21-13} = \frac{84}{8} = \frac{21}{2}$.

(C) We know that $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Substituting these into the expression:
$\frac{r_1-r}{a} + \frac{r_2-r}{b} = \frac{\frac{\Delta}{s-a} - \frac{\Delta}{s}}{a} + \frac{\frac{\Delta}{s-b} - \frac{\Delta}{s}}{b}$
$= \frac{\Delta}{a} \left( \frac{s - (s-a)}{s(s-a)} \right) + \frac{\Delta}{b} \left( \frac{s - (s-b)}{s(s-b)} \right)$
$= \frac{\Delta}{a} \left( \frac{a}{s(s-a)} \right) + \frac{\Delta}{b} \left( \frac{b}{s(s-b)} \right)$
$= \frac{\Delta}{s(s-a)} + \frac{\Delta}{s(s-b)}$
$= \frac{\Delta}{s} \left( \frac{1}{s-a} + \frac{1}{s-b} \right) = \frac{\Delta}{s} \left( \frac{s-b+s-a}{(s-a)(s-b)} \right)$
$= \frac{\Delta}{s} \left( \frac{2s-a-b}{(s-a)(s-b)} \right) = \frac{\Delta}{s} \left( \frac{c}{(s-a)(s-b)} \right)$
Since $\Delta^2 = s(s-a)(s-b)(s-c)$,we have $(s-a)(s-b) = \frac{\Delta^2}{s(s-c)}$.
Substituting this:
$= \frac{\Delta}{s} \cdot \frac{c \cdot s(s-c)}{\Delta^2} = \frac{c(s-c)}{\Delta} = \frac{c}{\frac{\Delta}{s-c}} = \frac{c}{r_3}$.

(C) Given,in $\triangle ABC$,$r=1$,$R=4$,and $\Delta=8$.
We know that $r = \frac{\Delta}{s}$,where $s$ is the semi-perimeter.
$1 = \frac{8}{s} \implies s = 8$.
Since $s = \frac{a+b+c}{2}$,we have $a+b+c = 2s = 16$.
We also know that $abc = 4R\Delta$.
$abc = 4 \times 4 \times 8 = 128$.
Now,we need to evaluate $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$.
$\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{c+a+b}{abc} = \frac{a+b+c}{abc}$.
Substituting the values,we get $\frac{16}{128} = \frac{1}{8}$.

(D) We know that $\sin 2C = 2 \sin C \cos C$ and $\sin 2B = 2 \sin B \cos B$.
Substituting these into the expression:
$\frac{1}{4}[b^2(2 \sin C \cos C) + c^2(2 \sin B \cos B)] = \frac{1}{2}[b^2 \sin C \cos C + c^2 \sin B \cos B]$.
Using the sine rule,$\sin C = \frac{c \sin A}{a}$ and $\sin B = \frac{b \sin A}{a}$,or simply using the projection formula $a = b \cos C + c \cos B$ and $\Delta = \frac{1}{2} bc \sin A$:
$\frac{1}{2}[b^2 \cos C \frac{2\Delta}{ab} + c^2 \cos B \frac{2\Delta}{ac}] = \frac{\Delta}{a}[b \cos C + c \cos B] = \frac{\Delta}{a}(a) = \Delta$.
Now,consider the expression $rr_1 \cot \frac{A}{2}$.
We know $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,and $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}} = \frac{s(s-a)}{\Delta}$.
Thus,$rr_1 \cot \frac{A}{2} = \frac{\Delta}{s} \cdot \frac{\Delta}{s-a} \cdot \frac{s(s-a)}{\Delta} = \Delta$.
Therefore,$\frac{1}{4}[b^2 \sin 2C + c^2 \sin 2B] = rr_1 \cot \frac{A}{2}$.