b Draw segments from Q and R to the angle bisector to get two triangles. Can you prove that these triangles are congruent?
A
aExample Solution:
B
b See solution.
Practice makes perfect
a Given the segment QR, a compass and straightedge can be used to draw its perpendicular bisector.
Place the compass' sharp end at one of the segment's endpoints. Draw an arc with a radius larger than half the distance between Q and R.
Keeping the compass length the same, draw a corresponding arc on the opposite side. The two arcs should now intersect at two distinct points.
The line that contains these two intersections can now be drawn using a straightedge.
This line is perpendicular to QR, and their intersection is at the midpoint of QR.
Finally, we will draw a segment from Q and R to an arbitrary point on the perpendicular bisector, P, to create △ PQR.
b According to the Perpendicular Bisector Theorem, if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. If we can show that the distance from Q and R to our arbitrary point P, are congruent, we know that △ PQR is an isosceles triangle.
We can prove that PQ≅ PR if we show that △ PQS≅ △ PRS. Notice that ∠ PSR and ∠ PSQ form a linear pair, and since ∠ PSR=90^(∘) it must be that ∠ PSQ=90^(∘) as well. We also notice that the triangles share PS as a side, which means it is congruent by the Reflexive Property of Congruence.
Now we can claim that △ PQS≅ △ PRS by the SAS (Side-Angle-Side) Congruence Theorem. Since PQ and PR are corresponding sides we know that they are congruent, which means P is equidistant from the endpoints of QR . Therefore, PQ≅ PR, which means △ PQR is an isosceles triangle.
