Find the segment $EM$ value in the figure below

Question

If $E$ is the excenter of triangle $ABC$ relative to $BC$, $AC=2(BC)=8$ and $AB=6$, calculate $EM$. Considering $AB \parallel EM$(Answer: $4.8$)

Here are the relations I found but couldn't finalize. I was not able to visualize the relationship between parallelism in the solution of the problem.

$S \triangle ABC =R(p-BC)=\sqrt{p(p-a)(p-b)p-c)}\\ p=\frac{8+6+4}{2} = 9\\ R =\frac{\sqrt{9(1)(3)(5)}}{(9-4)}=\frac{3\sqrt15}{5}\\ BO = p - AB = 9-6 = 3\\ BL = p - AC = 9-8 = 1\\ CH = p - AC = 9-8=1\\ \triangle ALI \sim \triangle AOE: \frac{r}{R} = \frac{AL}{AO} \implies r = \frac{3\sqrt{15}}{5}.\frac{5}{9} =\frac{\sqrt{15}}{3}$

$\triangle CEH: CE^2=1^2+R^2 \implies CE^2 = 1+\frac{15.9}{25} = \frac{160}{25}\therefore CE=\frac{4\sqrt10}{5}\\ \triangle EON: EN^2 = R^2+9 = 9+\frac{15.9}{25}=\frac{360}{25}\therefore EN = \frac{6\sqrt{10}}{ 5}\\ \triangle ACN \sim \triangle MCE \frac{EM}{AN} =\frac{CE}{CN}\implies EM = 12.\frac{\frac{4\sqrt{10}}{5}}{\frac{10\sqrt{10}}{5}}=4.8$

It remains to prove that ON = OB???

Answer 1

This is a good reference to recall some formulas: Incircle and excircles of a triangle.

The half-perimeter together with many things is calculated by OP. We have $p=s=9$.

The exradius $r_a=\sqrt{\frac{s(s-b)(s-c)}{(s-a)}}=\sqrt{\frac{9(9-8)(9-6)}{(9-4)}}=\frac{3\sqrt3}{\sqrt5}$.

By law of cosines, $\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{8^2+6^2-4^2}{2.8.6}=\frac78$ and thus $\sin A=\frac{\sqrt3\sqrt5}{8}.$

In the right triangle $\triangle EMH$, we have $\angle EMH=\angle A$ since $EM\parallel BA$ and $\angle EHA= 90^{\circ}$. So, $EM=\frac{r_a}{\sin A}=(\frac{3\sqrt3}{\sqrt5})(\frac{\sqrt3\sqrt5}{8})^{-1}=(\frac{3\sqrt3}{\sqrt5})(\frac{8}{\sqrt3\sqrt5})=\frac{24}{5}=4.8.$

Answer 2

Let’s add the point $D$, which marks the location at which $A$-excircle of $\triangle ABC$ touches its extended side $AB$. For brevity, we denote $\angle BAE$, the length of $EM$, and the semiperimeter of $\triangle ABC$ by $\theta$, $d$, and $s$ respectively. We also let $AB=c, BC=a$, and $CA=b$. We are aware of the fact that $BD=s-a$. which means that $AD =s$.

Since $\triangle EDA$ is a right-angled triangle, we shall write, $$AD=s=AE\cos\left(\theta\right) \qquad\rightarrow\qquad AE=\dfrac{s}{\cos\left(\theta\right)}.$$

Since $EM$ is parallel to $AB$, we have $\angle MEA=\angle DAE=\theta$. $AE$ is the angle bisector of $\angle BAC$ and, therefore, $\angle EAM = \angle DAE=\theta$. This makes $\triangle AME$ an isosceles triangle. Hence, we have, $$2d\cos\left(\theta\right) = AE\qquad\rightarrow\qquad d=\dfrac{s}{2\cos^2\left(\theta\right)}=\dfrac{s}{1+\cos\left(2\theta\right)}.\tag{1}$$

We apply the law of cosines to $\triangle ABC$ to obtain, $$\cos\left(2\theta\right)=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{7}{8}.\tag{2}$$

Semiperimeter of $\triangle ABC$ can be calcuted as, $$s=\dfrac{a+b+c}{2}=9\tag{3}$$

Now, we can find the value of $d$ by substituting the values obtained in (2) and (3) in (1) as shown below. $$d=\dfrac{9}{1+\dfrac{7}{8}}=\dfrac{72}{15}=4.8$$