Question

Triangle $ABC$ has side lengths $AB = 9$, $AC = 10$, and $BC = 17$. Let $X$ be the intersection of the angle bisector of $\angle A$ with side $\overline{BC}$, and let $Y$ be the foot of the perpendicular from $X$ to side $\overline{AC}$. Compute the length of $\overline{XY}$.

Answer 1

Answer:

$\dfrac{72}{19}$

Step-by-step explanation:

Consider triangle ABC. Segment AX is angle A bisector. Its length can be calculated using formula

$AX^2=\dfrac{AB\cdot AC}{(AB+AC)^2}\cdot ((AB+AC)^2-BC^2)$

Hence,

$AX^2=\dfrac{9\cdot 10}{(9+10)^2}\cdot ((9+10)^2-17^2)=\dfrac{90}{361}\cdot (361-289)=\dfrac{90}{361}\cdot 72=\dfrac{6480}{361}$

By the angle bisector theorem,

$\dfrac{AB}{AC}=\dfrac{BX}{XC}$

So,

$\dfrac{9}{10}=\dfrac{BX}{17-BX}\Rightarrow 153-9BX=10BX\\ \\19BX=153\\ \\BX=\dfrac{153}{19}$

and

$XC=17-\dfrac{153}{19}=\dfrac{170}{19}$

By the Pythagorean theorem for the right triangles AXY and CXY:

$AX^2=AY^2+XY^2\\ \\XC^2=XY^2+CY^2$

Thus,

$\dfrac{6480}{361}=XY^2+AY^2\\ \\\left(\dfrac{170}{19}\right)^2=XY^2+(10-AY)^2$

Subtract from the second equation the first one:

$\dfrac{28900}{361}-\dfrac{6480}{361}=(10-AY)^2-AY^2\\ \\\dfrac{22420}{361}=100-20AY+AY^2-AY^2\\ \\\dfrac{1180}{19}=100-20AY\\ \\20AY=100-\dfrac{1180}{19}=\dfrac{1900-1180}{19}=\dfrac{720}{19}\\ \\AY=\dfrac{36}{19}$

Hence,

$XY^2=\dfrac{6480}{361}-\left(\dfrac{36}{19}\right)^2=\dfrac{6480-1296}{361}=\dfrac{5184}{361}\\ \\XY=\dfrac{72}{19}$