In this assignment I want to look at the relationships between the medial triangle and the triangle constructed from the midpoints of the segments joining each vertex to the orthocenter of an acute triangle. Specifically, I want to prove that triangle LMN is similar to triangle ABC and congruent to the medial triangle DEF of the acute triangle.

Let's start by proving that triangle LMN is similar to triangle ABC. To do this we need to look at the triangles created by the altitudes. I need to show that triangle LHN similar to triangle AHC, triangle CHB similar to triangle NHM, and triangle BHA similar to triangle MHL to show that triangle LMN is similar to triangle ABC.

Triangle LHN similar to triangle AHC

Both triangles share angle AHC. We know that L is the midpoint of HA and N is the midpoint of HC which means sides LH and HN of triangle LHN are half the size of the sides of triangle AHC. Triangle AHC and triangle LHN are similar by SAS similarity. Therefore, side LN similar to side AC by a ratio of 1/2.

Triangle CHB similar to triangle NHM

Both triangles share angle CHB. We know that N is the midpoint of HC and M is the midpoint of HB which means sides HN and HM of triangle NHM are half the size of the sides of triangle CHB. Triangle CHB and triangle NHM are similar by SAS similarity. Therefore, side NM similar to side BC by a ratio of 1/2.

Triangle BHA similar to triangle MHL

Both triangles share angle BHA. We know that L is the midpoint of HA and M is the midpoint of HB which means sides HL and HM of triangle MHL are half the size of the sides of triangle BHA. Triangle BHA and triangle MHL are similar by SAS similarity. Therefore, side ML similar to side BA by a ratio of 1/2.

Showing triangle LMN is congruent to the medial triangle DEF is simple using what we know about the relationship between triangle ABC and triangle DEF. A medial triangle is similar to the original triangle by a ratio of 1/2.

Triangle DEF is similar to triangle ABC.

Proof:

claim 1: Triangle DBE is similar to triangle ABC.

The vertices D and E of triangle DEF are midpoints of sides AB and BC, respectively. So DB is 1/2 of AB and EB is 1/2 of CB. Both triangles share the angle ABC. So, triangle DBE is similar to triangle ABC by a ratio of 1/2 by SAS similarity. Therefore, DE is similar to AC by a ratio of 1/2.

claim 2: Triangle FEC is similar to triangle ABC.

The vertices F and E of triangle ECF are midpoints of sides AB and BC, respectively. So EC is 1/2 of BC and FC is 1/2 of AC. Both triangles share the angle BCA. So, triangle ECF is similar to triangle BCA by a ratio of 1/2 by SAS similarity. Therefore, EF is similar to AB by ratio of 1/2.

claim 3: Triangle ADF is similar to triangle ABC.

The vertices D and F of triangle ADF are midpoints of sides AB and AC, respectively. So AD is 1/2 of AB and AF is 1/2 of AC. Both triangles share the angle CAB. So, triangle ADF is similar to triangle ABC by a ratio of 1/2 by SAS similarity. Therefore, DF is similar to BC by a ratio of 1/2.

Since triangle DEF is similar to triangle ABC by a ratio of 1/2 and we showed that LMN is also similar to triangle ABC by a ratio of 1/2, then triangle DEF and triangle LMN must be congruent.