Angle bisector problem In triangle $ABC$, $AY$ is a perpendicular to the bisector $\angle ABC$ and $AX$ is a perpendicular to the bisector of $\angle ACB$. If $AB= 9cm$ , $AC=7cm$ and $BC= 4cm$, then the length of $XY$ is?
Any help would be much appreciated.
Thank you. 
 A: 
AX and AY, when both extended, cut the line BC at X’ and Y’ respectively.
Then, it is not that difficult to see that $\triangle BAX \cong \triangle BX’X$. Similarly, we have $\triangle CAY \cong \triangle CY’Y$. The consequences are:-
1) BX’ = BA, a known quantity and similarly, CY’ = CA; and
2) We can apply the midpoint theorem to $\triangle AX’Y’$ to get $XY = \dfrac {X’Y’}{2}$.
XY can then be found by observing that $X’Y’ = BX’ + CY’ – BC$.
A: Let $AB=c$, $AC=b$, $BC=a$, $\measuredangle BAC=\alpha$, $\measuredangle ABC=\beta$, $\measuredangle ACB=\gamma$, 
$R$ be a radius of circumcircle of $\Delta ABC$, $S$ be an area of the triangle
and let be our bisectors intersects in the point $O$.
Thus, $$\measuredangle XAY=90^{\circ}-\frac{\beta}{2}+90^{\circ}-\frac{\gamma}{2}-\alpha=180^{\circ}-90^{\circ}-\frac{\alpha}{2}=90^{\circ}-\frac{\alpha}{2}.$$
By law of sines for $\Delta AOC$ we obtain:
$$\frac{AO}{\sin\frac{\gamma}{2}}=\frac{b}{\sin\frac{\alpha+\gamma}{2}}$$ or
$$AO=\frac{b\sin\frac{\gamma}{2}}{\cos\frac{\beta}{2}}.$$
But $AO$ is a diameter of the circumcircle of $AXOY$, which says
$$XY=AO\sin\measuredangle XAY=\frac{b\sin\frac{\gamma}{2}\cos\frac{\alpha}{2}}{\cos\frac{\beta}{2}}=\frac{2R\sin\beta\sin\frac{\gamma}{2}\cos\frac{\alpha}{2}}{\cos\frac{\beta}{2}}=4R\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\cos\frac{\alpha}{2}.$$
Now, $p=\frac{a+b+c}{2}=\frac{4+7+9}{2}=10$, which says
$$S=\sqrt{10(10-4)(10-7)(10-9)}=6\sqrt5,$$
$$R=\frac{abc}{4S}=\frac{4\cdot7\cdot9}{4\cdot6\sqrt5}=\frac{21}{2\sqrt5}.$$
Now, by law of cosines for $\Delta ABC$ we obtain:
$$\cos\alpha=\frac{7^2+9^2-4^2}{2\cdot7\cdot9}=\frac{19}{21},$$
$$\cos\beta=\frac{4^2+9^2-7^2}{2\cdot4\cdot9}=\frac{2}{3}$$ and
$$\cos\gamma=\frac{4^2+7^2-9^2}{2\cdot4\cdot7}=-\frac{2}{7}.$$
Thus, $$\sin\frac{\beta}{2}=\sqrt{\frac{1-\frac{2}{3}}{2}}=\frac{1}{\sqrt6},$$
$$\sin\frac{\gamma}{2}=\sqrt{\frac{1+\frac{2}{7}}{2}}=\frac{3}{\sqrt{14}}$$ and
$$\cos\frac{\alpha}{2}=\sqrt{\frac{1+\frac{19}{21}}{2}}=2\sqrt{\frac{5}{21}}.$$
Id est,
$$XY=4\cdot\frac{21}{2\sqrt5}\cdot\frac{1}{\sqrt6}\cdot\frac{3}{\sqrt{14}}\cdot2\sqrt{\frac{5}{21}}=6.$$
Done!
A: Extend Ray AX and Ray BY meeting Line BC  at Q and P respectively.
Then AX  = XQ and AY = YP and ∆ ACQ and ∆ ABP are Isosceles. 
Hence, CQ = AC = 7 and BP = AB  = 9 and X is midpoint of AQ and Y is midpoint of AP. Also QP  = 12.
Hence, by Midpoint theorem, XY = 6. 
