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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

Internet references

Related pages in this website:

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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