In this question, angle A of a triangle is trisected, angle C is bisected, and BM is the triangle median.
|Triangle ABC is not obtuse.
|AM| = |MC|
m(BAP) = m(PAQ) = m(QAC)
m(ACN) = m(NCB)
|AQ| = 2
|BH| = ?
Note that AP and AQ are reflections about the angle bisector of A, and CP and CQ are coincident with the angle bisector of C (and thus reflections about it), so P and Q are Isogonal Conjugates. This means BP and BQ are reflections about the angle bisector of B, and m(MBC)=m(ABH).
If the median and altitude of triangle ABC are reflections about the bisector of B, as they are here, then ABC is a right triangle. (Proof) I can only imagine that the diagram is drawn this way to disguise the fact that B is a right angle.
To begin the solution, extend AP until it meets BC at P'. Now, from Ceva's Theorem,
(eq 1) CP' / P'B � BN / NA � AM / MC = 1, and then because
AM = MC,
(eq 2) CP' / P'B � BN / NA = 1
Let a = m(BAP), and note that since we're told that AQ=2, it follows that 2 cos a = AH.
Because triangles of equal height have area in proportion to the length of their bases,
(eq 3) AC AP' sin 2a / (AP' AB sin a) � CB CN sin c / (CN CA sin
c) = 1
(eq 4) 2 cos a / AB � CB = 1
(eq 5) AH / AB � CB = 1
Since ABC is a right triangle, AHB is similar to BHC, so AH/AB = BH/BC. From eq 5, replacing AH/AB with BH/BC, we get
(eq 6) BH / BC � CB = 1
(eq 7) BH = 1
Summary of geometrical theorems
Triangles -- a home page for anything about triangles.
Ceva's Theorem -- In triangle ABC, arbitrary cevians AD, BE and CF, meet at a point P iff AF/FB � BD/DC � CE/EA = 1
Isogonal Conjugates -- the relationship between the orthocenter and circumcenter of a triangle
Proof that the median and altitude of a right triangle are reflections about the bisector iff ABC is a right triangle.
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