Construct, with straightedge and compass, a ΔABM given |AB| and |BM| such that the altitude from vertex A, the median from vertex M, and the bisector of ∠ABM are concurrent.

Perhaps Bractals has given us a clue in the title...

Construction

Draw AB with given length and then a circle with radius |BM| and centre at B. Let this circle cut BA (produced if necessary) at P.
Now construct the mid-point, Q, of AP and draw a circle with centre at Q, and passing through B. M is either of the two points where the circles intersect.

In triangle ABM, the median MD and angle bisector BE can both be constructed. Then the third cevian, AC, can be drawn through their intersection and will be an altitude for the reasons given below:

Proof

Let |BC| = c,  |AB| = m  and  |BM| = |BP| = a,   so that |QB| = 0.5(a + m).

Since BE is the angle bisector, E divides AM in the ratio m : a.

Using Ceva’s Theorem in triangle ABM: (m/a)*([a - c]/c)*1 = 1,
which gives        m(a - c) = ac
                        ma = c(a + m)
                        m/c = (a + m)/a = 0.5(a + m)/(0.5a)
Thus:                |AB|/|BC| = |QB|/|BN|   where N is the mid-point of BM.

which shows that triangle ABC is similar to the right angled triangle QBN.

It follows that /ACB is a right angle and that AC is an altitude.

Posted by Harry on 2011-05-04 20:58:38