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= QZ
To prove that PY = QZ, we need to use the given information about the triangle and the properties of parallel lines.
First, let's draw triangle XYZ and label the given points and lines as described:
```
X
/ \
/ \
/ \
/ \
/_________\
Y Z
```
A is the midpoint of YZ.
B is the intersection of the bisector of the exterior angle at X and the extension of YZ.
The line through A parallel to XB intersects XY at P.
The line through A parallel to XB intersects XZ extended at Q.
We want to prove that PY = QZ.
To begin, let's examine the given relationship between A, B, and X.
Since A is the midpoint of YZ, we can conclude that A divides YZ into two equal parts, YA and AZ.
Now, let's consider the line through A parallel to XB.
Since XB is extended through B, we can use the property of parallel lines to conclude that angles XAP and XBY are alternate interior angles and therefore congruent.
Using the congruent angles XAP and XBY, we can apply the Alternate Interior Angles Theorem to conclude that XB is parallel to AP.
By the same reasoning, we can also conclude that XB is parallel to AQ.
Now, let's look at triangle XAP and triangle XAQ.
Since XB is parallel to AP and AQ, we have corresponding angles XPA and XQA that are congruent. Therefore, triangle XAP and triangle XAQ are similar by the Angle-Angle Similarity Theorem.
Since triangle XAP and triangle XAQ are similar, their corresponding sides are proportional.
Using that, we can say that PY / QZ = PA / AQ.
But, since A is the midpoint of YZ, we know that PA = AQ.
Therefore, we can conclude that PY / QZ = 1.
Thus, PY = QZ.
Therefore, we have proven that PY = QZ.