**If**$BY$ and $CZ$ bisect angles $\angle ABX$ and $\angle XCA$, respectively,

**then**$\angle ZXY$ is a right angle.

**Special Case**; Investigate

**the Converse**; and

**Ask a Related Question**.

**Special Case**: An easy way to simplify the problem is to assume that $\Delta ABC$ is isosceles. In this case, we can use special right triangles to prove our result.

$\Delta BZM$ is a 30-60-90 triangle, so $PX = ZM = BZ / 2$, and symmetry gives $PZ = ZY / 2 = BZ / 2.$ Thus $\Delta PZX$ is an isosceles right triangle, and, similarly, so is $\Delta PYX.$ Thus the result is proved.

**The Converse**makes the right angle the assumed and attempts to prove that two segments are angle bisectors. $\angle BAC = 120^{\circ}$, and $AX$ bisects $\angle BAC.$

**If**$\angle ZXY$ is a right angle,

**then**$BY$ and $CZ$ bisect angles $\angle ABX$ and $\angle XCA$, respectively.