Prove that the three Perpendicular Bisectors of the sides
of a triangle are concurrent.
(SubProof)
by
Summer Brown
We wish to prove that segment AC* is congruent to segment
BC*. We will use this congruence later in the main proof.

Since D is the midpoint of segment AC, we know that AD is
congruent to DC.

Also, since C*D is a perpendicular, we know that both angle
C*DA and C*DC are right angles. Right angles are always congruent,
so angle C*DA is congruent to angle C*DC.

C*D is congruent to itself by the identity property.
Hence, by the SideAngleSide (SAS) congruence
criterion, Triangle ADC* is congruent to Triangle CDC*. Since corresponding
parts of congruent triangles are congruent, AC* is congruent to CC*.
Similarly:

Since F is the midpoint of segment BC, we know that BF is
congruent to FC.

Also, since C*F is a perpendicular, we know that both angle
C*FB and C*FC are right angles. Right angles are always congruent,
so angle C*FB is congruent to angle C*FC.

C*F is congruent to itself by the identity property.
Hence, by the SideAngleSide (SAS) congruence
criterion, Triangle BFC* is congruent to Triangle CFC*. Since corresponding
parts of congruent triangles are congruent, BC* is congruent to CC*.
Since AC* is congruent to CC* and BC* is congruent to
CC*, AC* is congruent to BC* by transitivity. This
makes sense since AC*, BC*, and CC* are all radii of the circumcircle.
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