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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.
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.