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You need to use a compass.
We will trace line m and point P. Then we want to draw a line perpendicular to line m through point P by using a compass and a straightedge. First we will draw an arc with center at point P that intersects m at two points, A and B.
Next, we will draw an intersecting arc with center A. Using the same radius, we will draw another arc with center B. Then, we will label the intersection of the arcs Q.
Finally, we will draw a line through points P and Q. This line will be perpendicular to line m.
We want to show why the line passing through point P is perpendicular to line m. Therefore, we will first trace the segments AP, BP, AQ, and BQ.
Since AP and BP is drawn with the same compass width, AP is congruent to BP. Also, since we drew AQ and BQ is with the same compass width, AQ is congruent to BQ.
Notice that PQ is common to both △ APQ and △ BPQ. Since all sides of △ APQ and △ BPQ are congruent, we can conclude that △ APQ is congruent to △ BPQ by the Side-Side-Side Congruence Theorem. AP &≅ BP AQ &≅ BQ PQ &≅ PQ &⇓ △ APQ &≅ △ BPQ We know that corresponding parts of congruent triangles are congruent. Therefore, ∠ APQ and ∠ BPQ are congruent. Let's label the intersection point of two lines as C.
∠PCA= ∠PCB
Add terms
.LHS /2.=.RHS /2.