A proof that the lines of the three altitudes of a triangle are concurrent by Margaret Morgan Given triangle ABC, E is constructed to be the midpoint of AB and line EF is constructed be perpendicular to AB. Thus any point F is equidistant from A and B. Proof: Segment AE is congruent to Segment EB by construction. Segment EF is congruent to itself. Angles AEF and BEF are both right angles by construction. Thus, triangles AEF and BEF are congruent by SAS. This means BF is congruent to AF and F is equidistant from A and B. Next construct the perpendicular bisector of BC. H is the intersection of the two perpendicular bisectors (of AB and BC), so segments AH, BH, and HC are congruent. Since H is equidistant from points A and C, it lies on the perpendicular bisector of AC (line IH). Thus the perpendicular bisectors of triangle ABC are concurrent. Next, we construct line JL through B and parallel to AC, line KL through C and parallel to AB, and line JK through A and parallel to BC. Claim: Triangle ABC is congruent to triangles JBA, BCL, and ACK. Thus giving us that JA is congruent to AK, JB is congruent to BL and KC is congruent to CL. Meaning that A is the midpoint of JK, C is the midpoint of KL, and B is the midpoint of JL. Proof: Triangle JAB is congruent to Triangle ABC