I have the following statement that I have to prove. It is supposed to be a geometric proof. I have somewhat started the proof but I am stumped. Here is the statement.
Given any triangle, let A, B, C be the midpoints of the three sides of a triangle and let D be the base of one of the altitudes. Show that these 4 points are concyclic.
So far this is what I have…
Connect the points A, B and C to form a triangle. We know that any triangle has a circumscribed circle and so the points A, B and C all lie on the circumference of this circle. Now we want to show that D also lies on this circle. If X, Y, Z is our initial triangle, we note that AC // XZ, BC //XY and AB // YZ.
I do not know how to continue after this. Please help me!
Best Answer
$AHC$ is a right triangle and the median relative to hypotenuse is half the hypotenuse
$AH=\frac12 FD\to AH=AD$
$ADCB$ is a parallelogram so AD=BC$
therefore $AH=BC$
Furthermore $AB\parallel DE$ and this proves that $ABCH$ is an isosceles trapezoid which is always possible to inscribe in a circle.
Hope this helps
EDIT
The median $AH$ is half $DF$ because you can build a rectangle as I did in the second picture, diagonals are equal and bisect each other. $$...$$