When an altitude, bisector, and median make equal angles

Now that the year is over, I'd like to describe a solution to a challenging problem discussed in class.

Given: a triangle ABC, the altitude AD, the angle bisector AE, and the median AF divide the angle ∠BAC into four equal parts.

Prove: Prove that ∠BAC is a right angle.

Sketch of proof: There are several ways to solve this problem, but regardless of the method, it is a good idea to first start by drawing a figure involving all the given points and line segments:

We might be interested in investigating whether the sides satisfy Pythogoras's theorem, so it is a good idea to label lengths of line segments. So let the length AB = c, and the length AC = b. Next, one should label lengths of line segments as defined by the given. We start by setting the length BD = x. Since we are given that ∠BAD = ∠DAE, and that AD⊥BE, we can use ASA to ascertain that the triangles ADB and ADE are congruent. Hence the length DE = x. Furthermore, setting EF = y and using the fact that AE is the median, we get the length FC = 2x + y. Thus we arrive at the following figure:

Now consider the triangle AEC. AF is the bisector at A. Furthermore, recall the following.

Bisector proportionality theorem: An angle bisector in a triangle divides the opposite side in the ratio of the two sides.

Apply this theorem to our figure:

Thus conclude that ( y + 2x ) / b = y / c .

Consider the triangle ABC with bisector AE and apply the same theorem.

Conclude therefore that (2y + 2x) / b = 2x / c . We can get rid of the factor 2 on both sides and reduce to (y + x) / b =x / c .

Now add these two equations:

[ ( y + 2x ) / b = y / c ] + [ (y + x) / b = x / c ]

⇒

( y + 2x ) / b + (y + x) / b = y / c + x / c

⇒

( 2y + 3x ) / b = ( y + x ) / c

This equation will be useful below.

Next construct the point G which is the intersection of AC and the altitude of the triangle AEF from the vertex E. By an argument similar to the above using ASA surmise that AG is congruent to AE. Thus when considering the figure obtained by constructing the line EG and FG we get:

In particular, notice that in terms of lengths of line segments, the equation resolves to:

DC / AC = DF / AG.

That is, the lines AD and FG cut the transversals DC and AC proportionally. Thus using the converse of the transversal proportionality theorem, we can surmise that the lines AD and FG are parallel.

We are ready for the final part of the proof. Let the measured angle ∠BAD =∠DAE=∠EAF=∠FAC = θ. We are trying to prove that 4θ = 90°. Since ∠DFA is the complementary angle to ∠FAD in the right triangle ADF we have ∠DFA = 90 - 2θ. Furthermore, since the angles at H are right angles and EH = HG we can use SAS to surmise that the triangles EFH and GFH are congruent and therefore ∠EFA =∠GFA. Finally, since AD and FG are parallel ∠EFG is a right angle. Putting this altogether we have

∠EFA +∠GFA = ( 90 - 2θ )+ ( 90 - 2θ ) = 90

which becomes

180 - 4θ = 90

therefore

90 = 4θ

QED