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)

[![enter image description here](https://i.stack.imgur.com/42VDj.jpg)](https://i.stack.imgur.com/42VDj.jpg)

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

[![enter image description here](https://i.stack.imgur.com/BR7j4.jpg)](https://i.stack.imgur.com/BR7j4.jpg)

$D ~é ~excentro \therefore AD ~é~ bissetriz \triangle ABC\\
Teorema Bissetriz: \frac{BK}{AB}=\frac{KC}{AC}\rightarrow\frac{14-KC}{13}=\frac{KC}{15}\\
\therefore KC = 7,5 ~e~BK = 6,5\\
\triangle ABK\sim \triangle AML: Razão~Semelhança=\frac{13}{6,5} = 2\\
\therefore LM = \frac{6,5}{2}=3,25\\
De~forma~análoga: LN = 3,75\\
\triangle AHB: M(ponto~médio)\rightarrow HM = MB = 6,5\\
\triangle ACI:N(ponto ~médio) \rightarrow NI = NC = 7,5\\
\therefore \boxed{\color{red}x = 6,5+3,25+3,75+7,5 = 21 }$