In the straightedge and compass construction of the equilateral triangle above, the reasons can be used to prove that AC ≅ BC are:
A. AB and AC are radii of the same circle A, and AB and BC are radii of the same circle, so AB ≅ AC and AB ≅ BC, and AC ≅ BC
D. AB and AC are radii of the same circle and AB and BC are radii of the same circle, so AB ≅ AC and AB ≅ BC. AC and BC are both congruent to AB, so AC ≅ BC.
In Mathematics and Euclidean Geometry, an equilateral triangle can be defined as a special type of triangle that has equal side lengths and all of its three (3) interior angles are equal.
Since lines AB and AC are radii of the same circle and line AB and line BC are radii of the same circle, we can logically deduce that line AB would be congruent with line AC and line AB would be congruent with line BC.
This ultimately implies that line AC and line BC are both congruent to line AB, so based on the transitive property of equality, we have;
AC ≅ BC.