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

<JBA is congruent to <BAC by alternate interior angles
AB is congruent to itself
<ABC is congruent to <BAJ by alternate interior angles.

The triangles are congruent by ASA

Proof: Triangle BLC is congruent to Triangle ABC

<LBC is congruent to <BCA by alternate interior angles
BC is congruent to itself
<LCB is congruent to <CBA by alternate interior angles.

The triangles are congruent by ASA

Proof: Triangle KAB is congruent to Triangle ABC

<KAC is congruent to <BCA by alternate interior angles
AC is congruent to itself
<ACK is congruent to CAB by alternate interior angles.

The triangles are congruent by ASA

By the transitive property, all four triangles are congruent. Corresponding sides of congruent triangles are congruent, which proves the remainder of our claim.

If we construct the altitude of the triangle ABC through point A, it is also the perpendicular bisector of JK. It is the bisector of JK because it goes through the midpoint A. It is perpendicular to JK because it is constructed to be perpendicular to BC which is parallel to JK.

By similar reasoning, the altitude of ABC through the point C is the perpendicular bisector of KL and the altitude of ABC through point B is the perpendicular bisector of JL.

From the earlier proof, we know the perpendicular bisectors of a triangle are concurrent. Thus the altitudes of ABC are concurrent.