Find the size of the segment joining the foot of the perpendiculars of a scalene triangle The sides of a scalene triangle measure 13, 14, and 15 units. Two outer bisectors of different angles are drawn and the third vertex is drawn perpendicular to these bisectors. Calculate the size of the segment joining the foot of these perpendiculars.(Answer:21)

My progress ..I thought of using sine theorem, cosine theorem and Pythagoras but it will be very complicated. There is probably a simpler solution

$D ~é ~excentro \therefore AD ~é~ bissetriz \triangle ABC\\
Teorema Bissetriz: \frac{BK}{AB}=\frac{KC}{AC}\rightarrow\frac{14-KC}{13}=\frac{KC}{15}\\
\therefore KC = 7,5 ~e~BK = 6,5\\
\triangle ABK\sim \triangle AML: Razão~Semelhança=\frac{13}{6,5} = 2\\
\therefore LM = \frac{6,5}{2}=3,25\\
De~forma~análoga: LN = 3,75\\
\triangle AHB: M(ponto~médio)\rightarrow HM = MB = 6,5\\
\triangle ACI:N(ponto ~médio) \rightarrow NI = NC = 7,5\\
\therefore \boxed{\color{red}x = 6,5+3,25+3,75+7,5 = 21 }$
 A: 
As $D$ is intersection of external bisectors of $\angle B$ and $\angle C$, $AD$ must be internal bisector of $\angle A$.
So, $\angle ADB = 90^0 - \cfrac{\angle B}{2} - \cfrac{\angle A}{2} = \cfrac{\angle C}{2}$
Similarly, $\angle ADC = \cfrac{\angle B}{2}$
Now notice that quadrilateral $AIDH$ is cyclic.
So, $\angle AHI = \angle ADC = \cfrac{\angle B}{2}$
Also, $\angle BAH = \cfrac{\angle B}{2} = \angle AHI$. So $M$ is the midpoint of $AB$.
And similarly, it follows that $N$ is the midpoint of $AC$.
That leads to $HI = HM + MN + NI$
$= AM + \cfrac{BC}{2} + AN = 6.5 + 7 + 7.5 = 21$
A: The way you had started does lead to a solution, although it is a little calculative. Note that:
$$AH=13 \cos \left(\frac {\beta}{2}\right)$$
$$AI=15 \cos \left(\frac {\gamma}{2} \right)$$
Also, $\angle AHI=\alpha +\frac {\beta+\gamma}{2}=90°+\frac {\alpha}{2}$.
Now, let $HI=x$. From the cosine law on $\Delta HAI$, we have:
$$\cos (\angle AHI)=\cos\left(90°+\frac {\alpha}{2}\right)=-\sin \frac {\alpha}{2}=\frac {AH^2+HI^2-x^2}{2 AH \cdot AI}$$
Thus, $$x=\sqrt {AH^2+HI^2-2 AH \cdot AI \sin \frac {\alpha}{2}} {\tag 1}$$
Now, use the formulae:
$$\cos \frac {\alpha}{2}=\sqrt {\frac {s(s-a)}{bc}}$$
$$\cos \frac {\beta}{2}=\sqrt {\frac {s(s-b)}{ac}}$$
$$\cos \frac {\gamma}{2}=\sqrt {\frac {s(s-c)}{ab}}$$
Here $a=14$, $b=15$ and $c=13$, while $s=\frac {a+b+c}{2}=21$.
Hence, all the half-angle trig values are known, and can be substituted back in $(1)$ to obtain the answer.
