In triangle XYZ,A is the midpoint of YZ. The bisector of the exterior angle at X of XYZ cuts YZ produced at B. The line through A parallel to XB, cuts XY at P and XZ produced at Q.

1. Prove PX= QX

To prove that PX = QX, we need to show that the triangles APX and AQX are congruent.

In triangle XYZ:
Since A is the midpoint of YZ, we have AY = AZ.
The bisector of exterior angle X of XYZ cuts YZ produced at B, so BX is the bisector of angle YBZ. This implies that angle XBY = angle XBZ.

Now, let's look at the triangles APX and AQX:
AP is parallel to XB, so angle APX = angle XBY.
Since angle XBY = angle XBZ (as mentioned above), we have angle APX = angle XBZ.

AQ is parallel to XB, so angle AQX = angle XBZ.

Since angle APX = angle XBZ and angle AQX = angle XBZ, we have angle APX = angle AQX.

Also, since AY = AZ, we have AX = AX (common side).

By angle-side-angle (ASA) congruence, we can conclude that APX and AQX are congruent triangles.

And since congruent triangles have corresponding sides equal, we can say PX = QX.

Therefore, PX = QX is proven.