Properties of Triangles

The incentre of triangle formed by the points $\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right)$, $\left(x_3, y_3, z_3\right)$ is

$ \begin{aligned} & \left(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c},\frac{a z_1+b z_2+c z_3}{a+b+c}\right) \end{aligned} $

where $a, b, c$ are the sides of the triangle. The vertices are $A(0,0,0), B(3,0,0)$ and $C(0,4,0)$

Here, the sides are

$ A B=3, A C=4 $

and $B C=5$

The incentre of the triangle is

$ \begin{aligned} & \left(\frac{5(0)+4(3)+3(0)}{12}, \frac{5(0)+4(0)+3(4)}{12}, \frac{5(0)+4(0)+3(0)}{12}\right) \\ & =\left(\frac{12}{12}, \frac{12}{12}, 0\right)=(1,1,0) \end{aligned} $

We have, $r_1=4, r_2=8$ and $r_3=24$

$ \begin{aligned} \frac{1}{r} & =\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{4}+\frac{1}{8}+\frac{1}{24} \\ & =\frac{6+3+1}{24}=\frac{10}{24}=\frac{5}{12} \Rightarrow r=\frac{12}{5} \\ \Delta^2 & =r_1 \cdot r_2 \cdot r_3 \cdot r=4 \times 8 \times 24 \times \frac{12}{5} \end{aligned} $

So, $\Delta=\frac{96}{\sqrt{5}} \Rightarrow r_1=\frac{\Delta}{s-a}$

$\Rightarrow s-a=\frac{\Delta}{r_1}=\frac{96}{\sqrt{5} \times 4}=\frac{24}{\sqrt{5}}$

Similarly, $s-b=\frac{\Delta}{r_2}=\frac{96}{\sqrt{5} \times 8}=\frac{12}{\sqrt{5}} \cdots$

$ s-c=\frac{\Delta}{r_3}=\frac{96}{\sqrt{5} \times 24}=\frac{4}{\sqrt{5}} \ldots $

On adding Eqs. (i), (ii) and (iii), we get

$ \begin{aligned} & s-a+s-b+s-c=\frac{24}{\sqrt{5}}+\frac{12}{\sqrt{5}}+\frac{4}{\sqrt{5}} \\ & 3 s-(a+b+c)=\frac{40}{\sqrt{5}} \\ & 3 s-2 s=\frac{40}{\sqrt{5}} \Rightarrow s=\frac{40}{\sqrt{5}} \end{aligned} $

Then, $a=s-\frac{24}{\sqrt{5}}=\frac{40}{\sqrt{5}}-\frac{24}{\sqrt{5}}=\frac{16}{\sqrt{5}}$

Let $a, b, c$ be the sides of $\triangle A B C$, tha

$ \begin{gathered} \cot A=\frac{b^2+c^2-a^2}{4 \Delta} \\ \cot B=\frac{a^2+c^2-b^2}{4 \Delta}, \cot C=\frac{a^2+b^2-c^2}{4 \Delta} \end{gathered} $

So, $\Sigma \cot A=\cot A+\cot B+\cot C=\frac{1}{45}$

$ \begin{aligned} & {\left[b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-\right.} \\ & =\frac{a^2+b^2+c^2}{4 \Delta} \end{aligned} $

(B) In $\triangle A B C$,

$ \begin{aligned} & \cot \frac{A}{2}=\frac{s(s-a)}{\sqrt{s(s-a)(s-b)(s-c)}}=\frac{s(s-c)}{\Delta} \\ & \cot \frac{B}{2}=\frac{s-b}{\Delta} \\ & \cot \frac{c}{2}=\frac{s-c}{\Delta} \text { where, } s=\frac{a+b+c}{2} \\ & \text { Thus, } \begin{aligned} & \Sigma \cot \frac{A}{2}=\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{c}{2} \\ &=\frac{s(s-a)}{\Delta}+\frac{s(s-b)}{\Delta}+\frac{s(s-c)}{\Delta} \\ &=\frac{s}{\Delta}[s-a+s-b+s-c] \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{s}{\Delta}[3 s-(a+b+c)] \\ & =\frac{s}{\Delta}(3 s-2 s)=\frac{s^2}{\Delta}=\frac{(a+b+c)^2}{4 \Delta} \end{aligned} $

(C) We have, $\tan A: \tan B: \tan C=1: 2: 3$

So, $\tan A=k, \tan B=2 K, \tan C=3 K$

Now, $\tan C=\tan [\pi-(A+B)]=3 K$

$\begin{aligned} & \Rightarrow \quad-\tan (A+B)=3 K \\ & \Rightarrow \quad \tan (A+B)=-3 K \\ & \Rightarrow \frac{\tan A+\tan B}{1-\tan A \tan B}=-3 K \end{aligned}$

$ \Rightarrow \quad \frac{K+2 K}{1-2 K^2}=-3 K $

$ \Rightarrow \quad 1-2 K^2=-1 \Rightarrow 2 K^2=2 $

$ \Rightarrow \quad K^2=1 \Rightarrow K=1 $

$ \begin{aligned} & \sin A=\frac{1}{\sqrt{2}} \Rightarrow \sin B=\frac{2}{\sqrt{5}} \\ & \sin C=\frac{3}{\sqrt{10}} \end{aligned} $

$ \begin{aligned} &\begin{aligned} \sin A: \sin B & : \sin C \\ & =\frac{1}{\sqrt{2}}: \frac{2}{\sqrt{5}} \frac{: 3}{\sqrt{10}}=\sqrt{5}: 2 \sqrt{2}: 3 \end{aligned}\\ &\text { (D) We have, }\\ &\begin{aligned} & \cot \left(\frac{A}{2}\right): \cos \left(\frac{B}{2}\right): \cot \left(\frac{C}{2}\right)=3: 5: 7 \\ \Rightarrow & \frac{\cot A / 2}{3}=\frac{\cot B / 2}{7}=\frac{\cot C / 2}{9}=k \\ \Rightarrow \quad & \frac{s(s-a)}{3 \Delta}=\frac{s(s-b)}{7 \Delta}=\frac{s(s-c)}{9 \Delta} \\ \Rightarrow \quad & \frac{s-a}{3}=\frac{s-b}{7}=\frac{s-c}{9} \\ \Rightarrow \quad & \frac{s-b+s-c}{9+7}=\frac{s-a+s-c}{3+9} \\ \Rightarrow & =\frac{s-a+s-b}{3+7} \\ \Rightarrow & \frac{2 s-b}{16}=\frac{a}{16}=\frac{b}{12}=\frac{c}{10}=K \\ \Rightarrow & \quad a: b: c=16: 12: 10=8: 6: 5 \end{aligned} \end{aligned} $

$\begin{aligned} & r_1=2 r_2=3 r_3 \\ & \Rightarrow \frac{\Delta}{s-a}=\frac{2 \Delta}{s-b}=\frac{3 \Delta}{s-c} \\ & \Rightarrow \frac{1}{s-a}=\frac{2}{s-b}=\frac{3}{s-c}=k \text { (let) } \\ & \Rightarrow s-a=\frac{1}{k}, s-b=\frac{2}{k} \text { and } s-c=\frac{3}{k}, s=\frac{6}{k} \\ & \Rightarrow a=s-(s-a)=\frac{5}{k} \\ & \text { Similarly, } b=\frac{4}{k} \text { and } c=\frac{3}{k} \\ & \text { We know that } \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\ & \Rightarrow \frac{\sin A}{5}=\frac{\sin B}{4}=\frac{\sin C}{3} \\ & \Rightarrow \sin A: \sin B: \sin C=5: 4: 3\end{aligned}$

The circumradius $ R $ for a right-angled triangle is given by:

$ \frac{b}{\sin B} = 2R $

Since $\sin 90^\circ = 1$, we have:

$ b = 2R $

Therefore, $ R = \frac{b}{2} $.

The inradius $ r $ is given by:

$ r = (s-b) \tan \frac{B}{2} = (s-b) $

Since $\angle B = 90^\circ$, $\frac{B}{2} = 45^\circ$ and $\tan 45^\circ = 1$. Hence:

$ r = s-b $

Now, we compute $ 2(r + R) $:

$ 2(r + R) = 2 \left(s - b + \frac{b}{2}\right) = 2s - b $

Further simplifying, considering the semi-perimeter $ s = \frac{a+b+c}{2} $:

$ 2s = a + b + c $

Therefore, substituting back:

$ 2s - b = a + b + c - b = a + c $

So, the value is $ a + c $.

In a triangle $ \triangle ABC $, we have the relation:

$ (a-b)(s-c)=(b-c)(s-a) $

Rearranging the terms, we obtain:

$ (s-b)(s-c) + (s-b)(s-a) = 2(s-a)(s-c) $

Next, divide each term by $ (s-a)(s-b)(s-c) $ and multiply by the area of the triangle $ \Delta $:

$ \frac{\Delta}{s-a} + \frac{\Delta}{s-c} = \frac{2\Delta}{s-b} $

This simplifies to:

$ r_1 + r_3 = 2r_2 $

We have,

ABC is a triangle

$ \cos A+\cos B+\cos C $

$ \begin{aligned} & =2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+1-2 \sin ^2 \frac{C}{2} \\ & =1+2 \sin \frac{C}{2} \cos \left(\frac{A-B}{2}\right)-2 \sin ^2 \frac{C}{2} & {[\because A+B+C=\pi]} \\ & =1+2 \sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\sin \frac{C}{2}\right] \\ & =1+2 \sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)\right] \\ & =1+2 \sin \frac{C}{2}\left[2 \sin \frac{A}{2} \sin \frac{B}{2}\right] \\ & =1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \end{aligned} $

$ \begin{aligned} & =1+4\left[\sqrt{\frac{(s-b)(s-c)}{b c}} \sqrt{\frac{(s-a)(s-c)}{a c}}\right] \\ & =\frac{1+4(s-a)(s-b)(s-c)}{a b c} \\ & =\frac{1+4 s(s-a)(s-b)(s-c)}{s a b c} [\therefore \text { multiply and divide by } s] \\ & =1+4 \frac{\Delta^2}{s a b c}, \text { where } \Delta \text { is area of triangle } \\ & =1+\left(\frac{\Delta}{s}\right)\left(\frac{4 \Delta}{a b c}\right) \quad\left[\because r=\frac{\Delta}{s} \text { and } R=\frac{a b c}{4 \Delta}\right] \\ & =1+\frac{r}{R} \end{aligned} $

In triangle $ \triangle ABC $, given $ a = 26 $, $ b = 30 $, and $ \cos C = \frac{63}{65} $, we need to find the length of side $ c $.

We know the cosine rule for any triangle is given by:

$ \cos C = \frac{a^2 + b^2 - c^2}{2ab} $

Substituting the known values:

$ \frac{a^2 + b^2 - c^2}{2 \times a \times b} = \frac{63}{65} $

This implied expression becomes:

$ \frac{26^2 + 30^2 - c^2}{2 \times 26 \times 30} = \frac{63}{65} $

Now, solve this step-by-step:

Calculate $ 26^2 $ and $ 30^2 $:

$ 26^2 = 676, \quad 30^2 = 900 $

Substitute these into the equation:

$ \frac{676 + 900 - c^2}{1560} = \frac{63}{65} $

Simplify the numerator:

$ 676 + 900 = 1576 $

Substitute back:

$ \frac{1576 - c^2}{1560} = \frac{63}{65} $

Cross-multiply to solve for $ c^2 $:

$ 65(1576 - c^2) = 63 \times 1560 $

Compute $ 63 \times 1560 $:

$ 63 \times 1560 = 98280 $

Now, solve for $ c^2 $:

$ 65 \times (1576 - c^2) = 98280 $

Simplify the expressions:

$ 102440 - 65c^2 = 98280 $

Rearrange to find $ c^2 $:

$ 102440 - 98280 = 65c^2 $

Solve:

$ 4160 = 65c^2 $

Divide by 65 to isolate $ c^2 $:

$ c^2 = \frac{4160}{65} = 64 $

Take the square root:

$ c = \sqrt{64} = 8 $

Thus, the value of $ c $ is 8.

In $\triangle AFC$

$ \cos A=\frac{A F}{A C}=\frac{A F}{b} $

$ \Rightarrow \quad A F=b \cos A $

In $\triangle AFH$

$ \cos \left(90^{\circ}-2 B=\frac{A F}{A H} \Rightarrow \sin B=\frac{b \cos A}{x}\right. $

$ \Rightarrow \quad x=\left(\frac{b}{\sin B}\right) \cos A $

Using sine rule, we get

$ \begin{aligned} x & =\left(\frac{a}{\sin A}\right) \cos A \\ \Rightarrow \quad & \frac{a}{x}=\tan A \end{aligned}$

Similarly, $\frac{b}{y}=\tan B$ and $\frac{c}{z}=\tan c$

$ \begin{aligned} \therefore \quad \frac{a k c}{x y z} & =\tan A \tan B \tan C \\ & =\tan A+\tan B+\tan C \quad [\because A+B+C=\pi]\\ & =\frac{a}{x}+\frac{b}{y}+\frac{c}{z} \end{aligned} $

Given, in a $\triangle A B C$,

$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$

We know that,

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

Given,

$r_1=36, r_2=18 \text { and } r_3=12$

We know that, in a $\triangle A B C$

$\begin{aligned} \frac{1}{r} & =\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3} \\ \Rightarrow \frac{1}{r} & =\frac{1}{36}+\frac{1}{18}+\frac{1}{12} \Rightarrow \frac{1}{r}=\frac{2+4+6}{72} \\ r & =6 \end{aligned}$

Now, we know that,

$\begin{aligned} \Delta^2 & =r \cdot r_1 \cdot r_2 \cdot r_3=6 \times 36 \times 18 \times 12 \\ \Delta^2 & =6^2 \times 6^2 \times 6^2 \\ \Delta & =6 \times 6 \times 6=216 \end{aligned}$

Again we know that,

$\begin{aligned} r & =\frac{\Delta}{S}=\frac{216}{S} \\ \Rightarrow \quad & S=\frac{216}{6}=36 \end{aligned}$

In a $\triangle A B C, a=6$ units, $b=5$ units and $c=4$ units

As we know that,

$\begin{aligned} \cos A & =\frac{b^2+c^2-a^2}{2 b c} \\ \Rightarrow \cos A & =\frac{b^2+c^2-a^2}{2 b c}=\frac{5^2+4^2-6^2}{2 \times 5 \times 4}=\frac{1}{8} \end{aligned}$

As we know that,

$\begin{aligned} \cos 2 A & =\cos ^2 A-\sin ^2 A \\ & =2 \cos ^2 A-1=2 \times\left(\frac{1}{8}\right)^2-1 \\ & =\frac{1}{32}-1=\frac{-31}{32} \end{aligned}$

Within the context of triangle $ \triangle ABC $, the following relationship holds:

We know that:

$ \left(\tan \frac{A}{2} \cdot \tan \frac{B}{2} \cdot \tan \frac{C}{2}\right) \leq \frac{1}{3 \sqrt{3}} $

Thus, squaring both sides of the inequality, we get:

$ \begin{aligned} \left(\tan \frac{A}{2} \cdot \tan \frac{B}{2} \cdot \tan \frac{C}{2}\right)^2 & \leq \left(\frac{1}{3 \sqrt{3}}\right)^2 \\ & = \frac{1}{27} \end{aligned} $

Given that $A B C$ be triangle.

Using cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2 b c}$

$\cos B=\frac{c^2+a^2-b^2}{2 c a}, \cos C=\frac{a^2+b^2-c^2}{2 a b}$

Now, $2(b c \cos A+c a \cos B+a b \cos C)$

$=2\binom{b c \frac{b^2+c^2-a^2}{2 b c}+c a \frac{c^2+a^2-b^2}{2 c a}}{+a b \cdot \frac{a^2+b^2-c^2}{2 a b}}$

$=2\left(\frac{b^2+c^2-a^2}{2}+\frac{c^2+a^2-b^2}{2}+\frac{a^2+b^2-c^2}{2}\right)$

$=2\left(\frac{b^2+c^2+a^2}{2}\right)=a^2+b^2+c^2$

Given that $A B C$ is a triangle

So, $A+B+C=180^{\circ}$

Given, $C=60^{\circ}$

$\Rightarrow \quad A+B=120^{\circ} \quad \text{... (i)}$

We have by sine formula

$\begin{aligned} & \frac{\sin A}{a}=\frac{\sin B}{b} \\ \Rightarrow & \frac{\sin A}{\sin B}=\frac{a}{b}=2+\sqrt{3} \text { (given) } \end{aligned}$

Applying componendo and dividendo, we get

$\begin{aligned} & \frac{\sin A+\sin B}{\sin A-\sin B}=\frac{2-\sqrt{3}+1}{2+\sqrt{3}-1}=\frac{3+\sqrt{3}}{\sqrt{3}+1} \\ \Rightarrow & \frac{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)} \\ & =\frac{(3+\sqrt{3})(\sqrt{3}-1}{(\sqrt{3}+1)(\sqrt{3}-1)} \\ \Rightarrow & \frac{2\left(\frac{\sqrt{3}}{2}\right)}{2\left(\frac{1}{2}\right)} \cot \left(\frac{A-E}{2}\right)=\sqrt{3} \\ \Rightarrow & \sqrt{3} \cot \left(\frac{A-B}{2}\right)=\sqrt{3} \Rightarrow\left(\frac{A-B}{2}\right)=45^{\circ} \\ \Rightarrow & A-B=90 \quad \text{.... (ii)} \end{aligned}$

Solving Eqs. (i) and (ii), we get $\angle A=105^{\circ}$

Given, $a=2, b=3, c=4$

Thus,

$\begin{aligned} \cos C & =\frac{a^2+b^2-c^2}{2 a b} \\ & =\frac{4+9-16}{2 \times 2 \times 3}=\frac{-3}{12}=\frac{-1}{4} \end{aligned}$

$\begin{aligned} & (b+c) \cos A+(c+a) \cos B+(a+b) \cos C \\ & =b \cos A+c \cos A+c \cos B+a \cos B+a \cos C+b \cos C \\ & =(b \cos C+c \cos B)+(c \cos A+a \cos C) \\ & \quad \quad+(a \cos B+b \cos A) \quad \ldots \text { (i) } \end{aligned}$

Using projection formula,

$\begin{aligned} & a=(b \cos C+c \cos B) \\ & b=(c \cos A+a \cos C) \\ & c=(a \cos B+b \cos A) \end{aligned}$

Substituting above values in Eq. (i), we get

$a+b+c$

Thus, $(b+c) \cos A+(c+a) \cos B+(a+b) \cos C =a+b+c$

Let the coordinates of C be (x, y).

Slope of $A C \times$ Slope of $B C=-1$

$\begin{aligned} & \left(\frac{y-3}{x-2}\right) \times\left(\frac{y-5}{x-4}\right)=-1 \\ \Rightarrow & y^2-8 y+15=-x^2+6 x-8 \\ \Rightarrow & x^2+y^2-6 x-8 y+23=0 \quad \text{... (i)}\\ & (A C)^2=(B C)^2 \text{[ABC is an isosceles right angled triangle]} \end{aligned}$

$\begin{aligned} \Rightarrow & (x-2)^2+(y-3)^2=(x-4)^2+(y-5)^2 \\ \Rightarrow & x^2+4-4 x+y^2+9-6 y \\ & =x^2+16-8 x+y^2+25-10 y \\ \Rightarrow & 4 x+4 y=28 \Rightarrow x+y=7 \\ \Rightarrow & y=7-x\quad \text{... (ii)} \end{aligned}$

Substituting the value of $y$ from Eq. (ii) in the Eq. (i),

$\begin{aligned} & x^2+(7-x)^2-6 x-8(7-x)+23=0 \\ \Rightarrow & x^2+49+x^2-14 x-6 x-56+8 x+23=0 \\ \Rightarrow & 2 x^2-12 x+16=0 \Rightarrow x^2-6 x+8=0 \\ \Rightarrow & x=4,2 \end{aligned}$

If $x=4$, then $y=3$ and If $x=2$, then $y=5$. If $(x, y)=(4,3)$, then

Centroid of triangle is $\left(\frac{2+4+4}{3}, \frac{3+3+5}{3}\right) =\left(\frac{10}{3}, \frac{11}{3}\right)$

If $(x, y)=(2,5)$, then centroid of $\triangle A B C$ is

$\left(\frac{2+2+4}{3}, \frac{5+3+5}{3}\right)=\left(\frac{8}{3}, \frac{13}{3}\right)$

$\begin{aligned} & \text { } \text { In } \triangle A B C, a \neq b \\ & \frac{a \cdot \cos A-b \cos B}{a \cos B-b \cos A}+\cos C \\ & =\frac{a\left[\frac{b^2+c^2-a^2}{2 b c}\right]-b\left[\frac{a^2+c^2-b^2}{2 a c}\right]}{a\left[\frac{a^2+c^2-b^2}{2 a c}\right]-b\left[\frac{b^2+c^2-a^2}{2 b c}\right]}+\cos C \\ & =\frac{a^2\left(b^2+c^2-a^2\right)-b^2\left(a^2+c^2-b^2\right)}{a b\left(a^2+c^2-b^2\right)-a b\left(b^2+c^2-a^2\right)}+\cos C \\ & =\frac{a^2 b^2+a^2 c^2-a^4-b^2 a^2-b^2 c^2+b^4}{a^3 b+a b c^2-a b^3-a b^3-a b c^2+a^3 b} \\ & +\frac{a^2+b^2-c^2}{2 a b} \\ & =\frac{a^2 c^2-a^4-b^2 c^2+b^4}{2 a^3 b-2 a b^3}+\frac{a^2+b^2-c^2}{2 a b} \\ & =\frac{a^2 c^2-a^4-b^2 c^2+b^4}{2 a b\left(a^2-b^2\right)}+\frac{\left(a^2-b^2\right)\left(a^2+b^2-c^2\right)}{2 a b\left(a^2-b^2\right)} \\ & =\frac{a^2 c^2-a^4-b^2 c^2+b^4}{2 a b\left(a^2-b^2\right)} \\ \end{aligned}$

$+\frac{a^4+a^2 b^2-a^2 c^2-b^2 a^2-b^4+b^2 c^2}{2 a b\left(a^2-b^2\right)}$

$\begin{aligned} & a^2 c^2-a^4-b^2 c^2+b^4+a^4+a^2 b^2 \\ = & \frac{-a^2 c^2-b^2 a^2-b^4+b^2 c^2}{2 a b\left(a^2-b^2\right)} \\ = & \frac{0}{2 a b\left(a^2-b^2\right)}=0 \end{aligned}$

In a triangle $ \triangle ABC $, the given sides are $ a = 2 $, $ b = 3 $, and $ c = 4 $.

We start by calculating the semi-perimeter ($ s $) of the triangle:

$ s = \frac{a + b + c}{2} = \frac{2 + 3 + 4}{2} = \frac{9}{2} $

Next, we use the formula for $ \tan\left(\frac{A}{2}\right) $:

$ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} $

Substitute the values:

$ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{\left(\frac{9}{2} - 3\right)\left(\frac{9}{2} - 4\right)}{ \frac{9}{2}\left( \frac{9}{2} - 2 \right) }} = \sqrt{\frac{\left(\frac{9}{2} - \frac{6}{2}\right)\left(\frac{9}{2} - \frac{8}{2}\right)}{\frac{9}{2}\left(\frac{9}{2} - \frac{4}{2}\right)}} $

Simplify the fractions:

$ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{(9 - 6)(9 - 8)}{9(9 - 4)}} = \sqrt{\frac{3 \times 1}{9 \times 5}} $

Finally, we have:

$ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{1}{15}} $

The angles of $\triangle A B C$ are in ratio $1: 2: 3$.

Let $A=\theta, B=2 \theta, C=3 \theta$

$\begin{aligned} & \because \quad A+B+C=180^{\circ} \\ & \Rightarrow \quad \theta+2 \theta+3 \theta=180^{\circ} \\ & \Rightarrow \quad 6 \theta=180^{\circ} \\ & \Rightarrow \quad \theta=30^{\circ} \\ \end{aligned}$

$\begin{aligned} & \therefore A=30^{\circ}, B=60^{\circ} \text { and } C=90^{\circ} \\ & \because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \end{aligned}$

$\begin{aligned} & a=k \sin A, b=k \sin B \text { and } c=k \sin c \\ & \begin{aligned} a: b: c & =k \sin 30^{\circ}: k \sin 60^{\circ}: k \sin 90^{\circ} \\ & =\frac{1}{2}: \frac{\sqrt{3}}{2}: 1 \\ & =1: \sqrt{3}: 2 \end{aligned} \end{aligned}$

We know, $r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}$ and $r_3=\frac{\Delta}{s-c}$

Also, $\cot \left(\frac{A}{2}\right)=\frac{s(s-a)}{\Delta}, \cot \left(\frac{B}{2}\right)=\frac{s(s-b)}{\Delta}$

and $\cot \left(\frac{C}{2}\right)=\frac{s(s-c)}{\Delta}$

$\therefore r_1 \cot \left(\frac{A}{2}\right)+r_2 \cot \left(\frac{B}{2}\right)+r_3 \cot \left(\frac{C}{2}\right)$

$=\frac{\Delta}{(s-a)} \cdot \frac{s(s-a)}{\Delta}+\frac{\Delta}{(s-b)} \cdot \frac{s(s-b)}{\Delta}+\frac{\Delta}{(s-c)} \cdot \frac{s(s-c)}{\Delta}$

$\begin{aligned} & =s+s+s \\ & =3 s \end{aligned}$

$\begin{aligned} & \text { }(a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2} \\ & =\left(a^2+b^2\right)\left(\cos ^2 \frac{C}{2}+\sin ^2 \frac{C}{2}\right)+2 a b\left(\sin ^2 \frac{C}{2}-\cos ^2 \frac{C}{2}\right) \\ & =a^2+b^2+2 a b\left(1-2 \cos ^2 \frac{C}{2}\right) \\ & =a^2+b^2+2 a b(-\cos C) \\ & =a^2+b^2-2 a b \cos C \\ & =a^2+b^2-2 a b\left(\frac{a^2+b^2-c^2}{2 a b}\right) \\ & =a^2+b^2-a^2-b^2+c^2=c^2 \\ & \therefore(a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2}=c^2\end{aligned}$

$r_1=\frac{\Delta}{s-a}, r_2 \frac{\Delta}{s-b}$ and $r_3=\frac{\Delta}{s-c}$ ...... (i)

Also,

$\begin{aligned} \Delta^2 & =r r_1 r_2 r_3 \quad .....\text{(ii)}\\ \frac{1}{r} & =\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3} \quad .....\text{(iii)}\end{aligned}$

From Eq. (iii), we get

$\begin{aligned} & \frac{1}{r}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1 \\ \therefore \quad r & =1 \end{aligned}$

From Eq. (ii), we get $\Delta^2=(1)(2) (3) (6) =36$

$\Rightarrow \quad \Delta=6$

Also, $r=\frac{\Delta}{s}$

$\begin{aligned} & \Rightarrow \quad l=\frac{6}{s} \Rightarrow s=6 \\ & r_1=\frac{\Delta}{s-a} \\ & \Rightarrow s-a=\frac{\Delta}{r_1} \Rightarrow a=s-\frac{\Delta}{r_1} \\ & \Rightarrow \quad a=6-\frac{6}{2}=6-3 \\ & \Rightarrow \quad a=3 \Rightarrow r_2=\frac{\Delta}{s-b} \\ & \Rightarrow \quad b=s-\frac{\Delta}{r_2}=6-\frac{6}{3} \Rightarrow b=4 \\ & r_3=\frac{\Delta}{s-c} \\ & \Rightarrow \quad c=s-\frac{\Delta}{r_3}=6-\frac{6}{6}=6-1 \\ & \Rightarrow \quad c=5 \\ & \therefore(a, b, c)=(3,4,5) \\ & \end{aligned}$

$\begin{aligned} & a \tan A+b \tan B=(a+b) \tan \left(\frac{A+B}{2}\right) \\ & \Rightarrow a\left(\tan A-\tan \left(\frac{A+B}{2}\right)\right)=b\left(\tan \left(\frac{A+B}{2}\right)-\tan B\right) \\ & \Rightarrow a\left(\frac{\sin A}{\cos A}-\frac{\sin \left(\frac{A+B}{2}\right)}{\cos \left(\frac{A+B}{2}\right)}\right)=b\left(\frac{\sin \left(\frac{A+B}{2}\right)}{\cos \left(\frac{A+B}{2}\right)}-\frac{\sin B}{\cos B}\right) \\ & \Rightarrow a\left[\frac{\sin \left(A-\frac{(A+B)}{2}\right)}{\cos A \cos \left(\frac{A+B}{2}\right)}\right]=b\left[\frac{\sin \left(\frac{A+B}{2}-B\right)}{\cos B \cos \left(\frac{A+B}{2}\right)}\right] \\ & a\left[\frac{\sin \frac{(A-B)}{2}}{\cos A}\right]=b\left[\frac{\sin \frac{(A-B)}{2}}{\cos B}\right] \\ & \Rightarrow(k \sin A)\left[\frac{\sin \left(\frac{A-B}{2}\right)}{\cos A}\right]=(k \sin B)\left[\frac{\sin \left(\frac{A-B}{2}\right)}{\cos B}\right] \\ & \left\{\begin{array}{l}\operatorname{In} \triangle A B C \\ a=k \sin A \\ b=k \sin B\end{array}\right\} \\ & \Rightarrow \quad \tan A=\tan B \\ & \Rightarrow \quad \angle A=\angle B \Rightarrow A=B \\ & \end{aligned}$

Given : $AD=4;\angle DAB=\frac{\pi}{6};\angle ABE=\frac{\pi}{3}$

$\begin{aligned} A O=\frac{8}{3} \text { and } O D= & \frac{4}{3} \\\\ & {[\because O \text { divides } A D \text { in } 2: 1] } \end{aligned}$

Area of $\triangle A B C=2( \text{Area of} ~\triangle A B E)$

$\begin{aligned} & =2\left[\frac{3}{2}(\text { Area of } \triangle A O B)\right] \\\\ & =3 \times \text { area of } \triangle A O B \\\\ & =3 \times \frac{1}{2} \times B O \times A O \\\\ & =3 \cdot \frac{1}{2} \cdot \frac{8}{3 \sqrt{3}} \cdot \frac{8}{3} \\\\ & \qquad\left\{\because \tan \frac{\pi}{6}=\frac{B O}{A O} \Rightarrow B O=\frac{1}{\sqrt{3}} \times \frac{8}{3}\right\} \\\\ & =\frac{32}{3 \sqrt{3}} \end{aligned}$

$\begin{aligned} 2 \Delta^2\left(a^2+b^2+c^2\right) & =a^2 b^2 c^2 \\ a^2+b^2+c^2 & =\left(\frac{a b c}{\Delta}\right)^2 \cdot \frac{1}{2}=8 R^2 \end{aligned}$

$[\because a b c / \Delta=4 R$ where $R$-circumradius

$\begin{aligned} & \because \quad a=2 R \sin A, b=2 R \sin B, c=2 R \sin C] \\ & \Rightarrow 4 R^2\left(\sin ^2 A+\sin ^2 B+\sin ^2 C\right)=8 R^2 \\ & \Rightarrow \quad \sin ^2 A+\sin ^2 B+\sin ^2 C=2 \\ & \Rightarrow \quad(\cos 2 A+\cos 2 B+\cos 2 C)=-1 \\ & \Rightarrow \quad-1-4 \cos A \cos B \cos C=-1 \\ & \cos A \cos B \cos C=0 \\ & \end{aligned}$

So, any one among $A, B$ and $C$ has to be $\frac{\pi}{2}$.

So, right angled triangle.

$\because r=\frac{\Delta}{s}, r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}$,

$r_3=\frac{\Delta}{s-c} \text { and } 2 s=a+b+c$

Then, $r_1+r_2+r_3-r=\frac{\Delta}{s-a}+\frac{\Delta}{s-b}+\frac{\Delta}{s-c}-\frac{\Delta}{s}$

$\begin{aligned} & =\Delta\left(\frac{2 s-(a+b)}{(s-a)(s-b)}\right)+\Delta\left(\frac{c}{s(s-c)}\right) \\ & =\Delta \frac{(2 s-2 s+c)}{(s-a)(s-b)}+\Delta\left(\frac{c}{s(s-c)}\right) \\ & =\Delta\left[\frac{c s(s-c)+c(s-a)(s-b)}{s(s-a)(s-b)(s-c)}\right] \\ & \quad\left[\because \Delta^2=s(s-a)(s-b)(s-c)\right] \\ & =\frac{\Delta c}{\Delta^2}(s(s-c)+(s-a)(s-b)) \\ & \left.=\frac{c}{\Delta}\left(s^2-s c+s^2-(a+b) s+a b\right)\right) \end{aligned}$

$\begin{aligned} & =\frac{c}{\Delta}\left(2 s^2-(a+b+c) s+a b\right) \\ & =\frac{c}{\Delta}\left(2 s^2-2 s^2+a b\right)[\because 2 s=a+b+c] \\ & =\frac{a b c}{\Delta}=4\left(\frac{a b c}{4 \Delta}\right)=4 R\end{aligned}$

In $\Delta ABC,a=3,b=4,\sin A=3/4$

$\therefore$ $\cos A = \sqrt {1 - {{\sin }^2}A} = \sqrt {1 - {9 \over {16}}} = {{\sqrt 7 } \over 4}$

But $\cos A = {{{b^2} + {c^2} - {a^2}} \over {2bc}}$

$ \Rightarrow {{\sqrt 7 } \over 4} = {{{4^2} + {c^2} - {3^3}} \over {2\,.\,4\,.\,c}}$

$\begin{array}{lrlrl}\Rightarrow 2 \sqrt{7} c =7+c^2 \\ \Rightarrow c^2-2 \sqrt{7} c+7 =0 \\ \Rightarrow (c-\sqrt{7})^2 =0 \\ \therefore c =\sqrt{7} \\ \text { Now, } \cos B =\frac{c^2+a^2-b^2}{2 a c} \\ \Rightarrow \cos B =\frac{7+9-16}{2 \cdot 3 \cdot \sqrt{7}}=0 \Rightarrow B=90 \Upsilon \\ \therefore \angle C B A =90 \Upsilon\end{array}$

$A=75 Y, B=45 \Upsilon$, then $C=180 Y-(75 Y+45 Y)=60 \Upsilon$

$\because \quad \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{C}{\sin C}=2 R$ where $R$ is circumradius

$\Rightarrow \quad a=2 R \sin A, b=2 R \sin B, c=2 R \sin C$

Then, $b+c \sqrt{2}=2 R(\sin B+\sqrt{2} \sin C)$

$\begin{aligned} & =2 R(\sin 45 Y+\sqrt{2} \sin 60 Y) \\ & =2 R\left(\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}\right)=2 R\left(\frac{\sqrt{3}+1}{\sqrt{2}}\right) \\ & =2 \cdot 2 R\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right) \\ & =2 \cdot(2 R \sin 75 Y) \\ & =2(2 R \sin A) \\ & =2 a \end{aligned}$

In $\triangle A B C$

$\begin{aligned} & a=2 R \sin A=4 R \sin \frac{A}{2} \cos \frac{A}{2} \\ & b=2 R \sin B=4 R \sin \frac{B}{2} \cos \frac{B}{2} \\ & r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\ & r_1=4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \\ & r_2=4 R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2} \\ & r_3=4 R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} \end{aligned}$

$\text { Now, } \begin{aligned} & a b-r_1 r_2=16 R^2 \sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{A}{2} \\ & \cos \frac{B}{2}\left(1-\cos ^2 \frac{C}{2}\right) \\ &= 16 R^2 \sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{B}{2} \sin ^2 \frac{C}{2} \end{aligned}$

$\therefore \quad \frac{a b-r_1 r_2}{r_3}=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=r$

Let $A, B, C$ is represented as $\mathbf{a}, \mathbf{b}, \mathbf{c}$ respectively.

Mid-points $D, E, F$ is $\frac{\mathbf{a}+\mathbf{b}}{2}, \frac{\mathbf{a}+\mathbf{c}}{2}, \frac{\mathbf{b}+\mathbf{c}}{2}$ respectively.

$\begin{aligned} \text { Then, } \mathbf{B E} & =\frac{\mathbf{a}-2 \mathbf{b}+\mathbf{c}}{2}, \mathbf{A F}=\frac{\mathbf{b}-2 \mathbf{a}+\mathbf{c}}{2}, \\ \mathbf{C D} & =\frac{\mathbf{a}+\mathbf{b}-2 \mathbf{c}}{2} \\ \mathbf{B E}+\mathbf{A F} & =\frac{2 \mathbf{c}-\mathbf{a}-\mathbf{b}}{2}=\frac{-(\mathbf{a}+\mathbf{b}-2 \mathbf{c})}{2} \\ & =-\mathbf{C D}=\mathbf{D C} \end{aligned}$