Length Of An Angle Bisector?

Relevant wiki: Area of Triangles - Problem Solving - Medium

Let the sides opposite to angle $A,B , C$ be $a,b,c$ respectively while $\angle BAC=\theta$ so that $\angle BAP=\angle CAP=\dfrac{\theta}{2}$ and $AP=x$ .

$ar(\triangle ABC)=ar(\triangle ABP)+ar(\triangle ACP)$

$\implies \dfrac{1}{2}bc\sin \theta=\dfrac{1}2cx\sin\dfrac{\theta}2+\dfrac{1}2bx\sin\dfrac{\theta}{2}$ $(Since \small{\color{#D61F06}{\text{ area}=\dfrac{1}{2}ab\times(\text{ Included angle}}})$

Use double angle formula and then reaarange for $x$ to get:-

$\Large \boxed{x=\dfrac{2bc}{b+c}\cos\dfrac{\theta}2}$

Special case: When $\theta=90°$ , $x=\dfrac{2bc}{b+c}\cos (45°)$

For the given question $b=5,c=12$ , Hence:

$x=\dfrac{2\cdot 5\cdot 12}{5+12}\cos (45°)=\dfrac{60\sqrt 2}{17}$

$\therefore 60+2+17=\large\color{#3D99F6}{79}$

Relevant wiki: Area of Triangles - Problem Solving - Medium

We can use Stewart's Theorem to generalise this for any triangle. First, I will prove Stewart's Theorem, which states $a(p^2+mn)=b^2m+c^2n$ for any cevian of a triangle (Below is the labeling definitions)

Let $\Delta ABC$ have sides $a$ , $b$ and $c$ as usual. Let $AP$ be a cevian, with $P$ on $a$ , with length $p$ . Let $P$ divide $BC$ into lengths $BP=m$ and $CP=n$ .

First, we will find an expression for $\cos \angle APB$ . But, by Cosine Rule, this is simply $\cos \angle APB = \dfrac {p^2+m^2-c^2}{2pm}$ . Similarly, $\cos \angle APC = \dfrac {p^2+n^2-b^2}{2pn}$ . But, $\angle APB$ and $\angle APC$ are supplementary, so their cosines add to 0. Thus, we have:

$\begin{aligned} \dfrac {p^2+m^2-c^2}{2pm}+\dfrac {p^2+n^2-b^2}{2pn} &= 0\\ p^2n+m^2n-c^2n+p^2m+n^2m-b^2m&=0\\ p^2m+p^2n+m^2n+mn^2&=b^2m+c^2n\\ (m+n)(p^2+mn)&=b^2m+c^2n\\ a(p^2+mn)&=b^2m+c^2n \end{aligned}$

Thus, we have proven Stewart's Theorem.

Now, by Angle Bisector Theorem, $\dfrac {c}{b}=\dfrac{m}{n}$ . Let $m=kc$ , $n=kb$ , with $k=\dfrac{a}{b+c}$ . Now, we can substitute this into the above expression to get

$\begin{aligned} a(p^2+k^2bc)&=kb^2c+kc^2b\\ p^2&=\dfrac{kbc(b+c-ka)}{a}\\ &= bc\left ( \dfrac {a}{b+c} \right ) \left ( \dfrac {b+c-\frac{a^2}{b+c}}{a} \right )\\ &= bc\left ( \dfrac {a}{b+c} \right ) \left ( \dfrac {(b+c)^2-a^2}{a(b+c)} \right )\\ &= bc \left (\dfrac {(b+c)^2-a^2}{(b+c)^2} \right )\\ &= bc (1- \left (\dfrac {a}{b+c} \right )^2)\\ p &= \sqrt{bc \left (1- \left ( \dfrac {a}{b+c} \right ) ^2 \right )} \end{aligned}$

Thus, the length of the angle bisector of a general triangle is $\sqrt{bc \left (1- \left (\dfrac {a}{b+c} \right )^2 \right )}$ . If the angle being bisected is a right angle, we can substitute $a=\sqrt{b^2+c^2}$ to get that the length is $\dfrac {bc\sqrt{2}}{b+c}$ . Thus, if we substitue $b=5, c=12$ , we get the length being $\dfrac {60\sqrt{2}}{17}$ . Thus, the answer for this problem is $60+2+17=\boxed{79}$ .

$Applying \ angle\ bisector\ theorem, AM=\dfrac{5*13}{5+12}=\dfrac{5*13}{17}.\\ CosMAB=\dfrac 5 {13}.\\ Applying \ Cos\ Law\ to\ \Delta\ MAB,\\ BM^2=AM^2+AB^2 - 2*AM*AB*CosMAB=\left(\dfrac{5*13}{17}\right)^2 + 25 - 2*\dfrac{5*13}{17}*5*\dfrac 5 {13}\\ \scriptsize\ \ \ \ \ \ =25\left\{\left(\dfrac{13}{17}\right)^2+1 - 2*\left(\dfrac{5*13}{17}\right)\right\}=\dfrac{25}{17^2}*(13^2 + 17^2 - 170) \scriptsize\ \ \ \ \ =\dfrac{25}{17^2}*2*(12^2)=\dfrac{(5^2*12^2)*2}{17^2}=\dfrac{a^2*b}{c^2}\\ \implies\ \ a+b+c=60+2+17=\color{#D61F06}{79} \\ \ \ \ \\$ $OR$ $\color{#3D99F6}{\text{On second though, late night, the simple solution is as under. }\\ SinA=\frac {12}{13},\ \ CosA=\frac 5 {13},\ \ SinA + CosA=\dfrac {17}{13}.\\ SinAMB=Sin\{MBA + BAM\}=Sin\{45 + BAM\}=\dfrac 1 {\sqrt2}*\{SinA + CosA\}=\dfrac 1 {\sqrt2}*\dfrac {17}{13}.\\ Using\ Sin\ Law\ in\ \Delta\ AMB,\\ BM=SinA*\dfrac{AB}{SinAMB}=\dfrac {12}{\color{#D61F06}{13}}*5*\dfrac{\sqrt2*\color{#D61F06}{13}}{17}=\dfrac{12*5*\sqrt2}{17}=\dfrac{a*\sqrt b} c.\\ a+b+c=60+2+17=79. }$

solution

We can use Sine rule in all three triangles.

In $\triangle ABC$ .

$\Rightarrow \dfrac{BC}{\sin A}=\dfrac{AC}{\sin B}=\dfrac{AB}{\sin C}$

$\dfrac{5}{\sin A}=\dfrac{13}{\sin 90°}=\dfrac{12}{\sin C}$

$\boxed{ \text{Sin A}=\dfrac{5}{13}}$ $\boxed{\sin C=\dfrac{12}{13}}$

Now, in $\triangle ABD$ .

$\Rightarrow \dfrac{\color{#3D99F6}{BD}}{\sin A}=\dfrac{AD}{\sin B}=\dfrac{AB}{\sin D}$

$\dfrac{\color{#3D99F6}{BD}}{\frac{5}{13}}=\dfrac{13-\color{#EC7300}{x}}{\sin 45°}$

$\dfrac{13\color{#3D99F6}{BD}}{5}=\sqrt{2}(13-\color{#EC7300}{x})$ .... $(1)$

Now, in $\triangle BDC$ .

$\Rightarrow \dfrac{\color{#3D99F6}{BD}}{\sin C}=\dfrac{DC}{\sin B}=\dfrac{BC}{\sin D}$

$\dfrac{\color{#3D99F6}{BD}}{\frac{12}{13}}=\dfrac{\color{#EC7300}{x}}{\sin 45°}$

$\dfrac{13\color{#3D99F6}{BD}}{12}=\sqrt{2}\color{#EC7300}{x}$ .... $(2)$

From $(1)$ and $(2)$ .

$\boxed{\color{#EC7300}{x}=\dfrac{65}{17}}$

Pluging the value of $\color{#EC7300}{x}$ in $(2)$ .

$\Rightarrow \dfrac{13\color{#3D99F6}{BD}}{12}=\sqrt{2}×\dfrac{65}{17}$

$\boxed{\color{#3D99F6}{BD}=\dfrac{60\sqrt{2}}{17}}$

$\dfrac{60\sqrt{2}}{17}=\dfrac{a\sqrt{b}}{c}$

$\therefore a+b+c=60+2+17=\boxed{\color{#D61F06}{79}}$

Let $d$ be an angle bisector. Applying Stewart's theorem, $d=\sqrt { \frac { { c }^{ 2 }n+{ b }^{ 2 }m-nma }{ a } }$ . Appyling angle bisector theorem, $\frac { n }{ b } =\frac { m }{ c }$ and, as $m+n=a$ , we can solve for $m$ and $n$ , and find the value of the length of the angle bisector.

In the given particular case, $m=65/17$ and $n=156/17$ , then $d=\frac { 60\sqrt { 2 } }{ 17 }$

I simply solved this problem by imagining the triangle on an x-y coordinate plane (with the right angle at the origin). In this case, the angle bisector could be represented by the equation x = y and the hypotenuse of the triangle could be represented by the equation 12x + 5y = 60 (or 5x + 12y = 60; order doesn't really matter in this case). Solving the system of equations, we get that the point of intersection is (60/17, 60/17), which is 60 sqrt(2) / 17 units away from the origin. Adding these up, we get 60 + 17 + 2 = 79. Is there anyone else who did this?

Relevant wiki: Angle Bisector Theorem

Let $AB=a, BC=b, AC=c, DC=x$ and $BD=y$

By angle bisector theorem

$\frac{a}{b}=\frac{c-x}{x} \Rightarrow$ $ax=bc-bx$ so $x=\frac{bc}{a+b}$

Then by law of sines

$\frac{\frac{bc}{a+b}}{ \sin 45°}=\frac{y}{ \sin \angle C}$

$y=\frac{\frac{bc}{a+b}}{ \sin 45°}\times \sin \angle C$

$y=\frac{\frac{bc}{a+b}}{ \sin 45°}\times \frac{a}{c}$

$\boxed{y=\frac{ab}{(a+b) \sin45°}}$

This is the bisector's length of the angle bisector in any rectangle triangle given only their cathetus.

Making a substitution we get angle bisector $=\frac{60\sqrt{2}}{17}$ with $\boxed{a+b+c=79}$

Similar problem: Cool Bisector

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