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.PY/PX = ZA/ BA = QZ/QX
To prove the given statement, we can use the concept of similar triangles and parallel lines.
First, let's establish some triangle congruence. Since A is the midpoint of YZ, we have AY = AZ. Also, since the bisector of the exterior angle at X cuts YZ extended at B, we have ∠XBA = ∠XBY. Combining these two facts, we can conclude that triangles XAY and XAZ are congruent by the SAS (Side-Angle-Side) congruence theorem.
Now, let's consider the lines AP and XB. These lines are parallel since they are both cut by transversal line XY. Thus, we can conclude that triangles PXY and BXY are similar by the corresponding angles postulate.
Using the concept of similar triangles, we have PY/PX = BY/BX. (1)
Similarly, considering the lines XQ and AB, we can conclude that triangles XBZ and XBA are similar. Therefore, we have QZ/QX = BA/BX. (2)
From (1), we know PY/PX = BY/BX, and from (2), we know QZ/QX = BA/BX.
Since we established that triangles XAY and XAZ are congruent, we have BY = BZ. Let's substitute this in equation (1):
PY/PX = BY/BX = BZ/BX
Since YZ extended through B results in BA (BY = BZ), we can replace BZ and BA in the above equation:
PY/PX = BZ/BX = BA/BX
This confirms that PY/PX = BA/BX.
Finally, using the concept of alternate interior angles (because AP is parallel to XB), we have ∠PXQ = ∠BAX. Therefore, triangles QXZ and PAX are similar. Using the concept of similar triangles, we can also conclude that ZA/QZ = AX/PX.
Combining this with the previous result, we have:
PY/PX = ZA/BA = QZ/QX
Thus, we have proven that PY/PX = ZA/BA = QZ/QX in triangle XYZ.