# In a right-angled triangle of equal opposite and adjacent sides, why is the intersection of the median lines one third of the triangles height?

In a right-angled triangle of equal opposite and adjacent lengths, I was told that when you drew a line from the median of the hypotenuse to its opposite vertex, and then drew another line from one other side to its respective opposite vertex, the point of intersection of the lines was one third the height of the triangle. Is this true? And if so, why?

I believe the picture speaks for itself. The simple answer is that in an isosceles right triangle the median to the hypotenuse is also an altitude. But in for a penny, in for a pound.

Let $\bigtriangleup$ABC be an isosceles right triangle with C the vertex opposite the hypotenuse. Let D be the midpoint of the hypotenuse AB, E the midpoint of BC and F the point of intersection of the segments CD and AE. We want to show that CD is an altitude and that CD = 3FD.

AC = BC, AD = BD and $\angle$ ABC = $\angle$ CAB, so $\bigtriangleup$ BCD is congruent to $\bigtriangleup$ ACD. So $\angle$ ADC = $\angle$ BDC. $\angle$ ADC and $\angle$ BDC are supplementary and equal so they are right angles. CD is therefore an altitude.

Construct an auxiliary segment DE.

$\bigtriangleup$ DBE and $\bigtriangleup$ ABC share $\angle$ DBE. BA = 2BD. BC=2BE. So $\bigtriangleup$ DBE and $\bigtriangleup$ ABC are similar.

That means AC = 2DE and $\angle$ BED = $\angle$ BCA, which in turn means DE is parallel to AC. So $\angle$ ACF = $\angle$ EDF. $\angle$ CAF = $\angle$ DEF. So $\bigtriangleup$ ACF is similar to $\bigtriangleup$ DEF and CF = 2FD.

Therefore, CD = 3FD and we're done.