Assignment 4
| Start with a triangle ABC. Construct the midpoint of BC. Label the midpoint M. |  |
| Construct the perpendicular bisector (x) of BC. |  |
| Construct any point (D) on x and construct DB and DC. |  |
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Since BM=CM, angle(BMD)=angle(CMD), and DM=DM, Triangle(BMD) |
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| Now, construct the perpendicular bisector (y) of AC. Since AC and BC are not parallel, lines x and y must intersect. |  |
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Merge point D to the point of intersection for lines x and y. AF=FC, angle (AFD)=angle(CFD), and DF=DF. Triangle (AFD) Therefore AD=DC. By the Axiom that states that every segement is congruent to itself, we know that if AD=DC and BD=DC, then AD=BD. |
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| Therefore, D is on the perpendicular bisector of AB as wel, yielding that the perpendicular bisectors of the sides of a triangle are concurrent. |  |
Triangle(CMD)
by the Side-Angle-Side Theorem. Therefore, BD=DC.