Given: Triangle ABC
Altitudes AE, BF, CD (where each of E,F,D is the foot of that altitude)
IH parallel to side AB passing through C
as shown in the sketch below.
Proof:
Segment GA is part of segment GI, which is parallel to segment BC by hypothesis. So GA is parallel to BC. Similarly, segment GB is part of segment GH, and therefore GB is parallel to segment AC, as a result of the hypothesis. Therefore, quadrilateral GBCA has two pairs of parallel sides, so GBCA is a parallelogram. Then since GA and BC are opposite sides of a parallelogram, GA is congruent to BC.
Segment AI is part of segment GI, which is parallel to segment BC by hypothesis. So AI is parallel to BC. Similarly, segment CI is part of segment HI, and therefore CI is parallel to segment AB, as a result of the hypothesis. Therefore, quadrilateral ABCI has two pairs of parallel sides, so ABCI is a parallelogram. Then since AI and BC are opposite sides of a parallelogram, AI is congruent to BC.
Then we know that GA is congruent to AI since we have shown both congruent to BC.
Since AE is the altitude to BC, angles AEB and AEC are both right angles. GI is parallel to BC, and A lies on GI. So GAE and AEC, as well as IAE and AEB, are pairs of alternate interior angles. Then GAE is congruent to AEC and IAE is congruent to AEB. So GAE and IAE are right angles. Therefore, line AE is perpendicular to GI, and we know that GA is congruent to A, so line AE is the perpendicular bisector of GI.
Symmetrical arguments show that line BF is the perpendicular bisector of GH (using quadrilaterals HCAB and BCAG) and line CD is the perpendicular bisector of HI (using quadrilaterals HBAC and CBAI).
To summarize, lines AE, BF, and CD are the perpendicular bisectors of the sides of triangle GIH.
But we have the following result proved in Assignment 8, Problem 12:
Lemma: The three perpendicular bisectors of the sides of a triangle are concurrent
Then, by the lemma AE, BF, and CD are concurrent.
