Solution of Ricardo Barroso Campos Universidad de Sevilla

Solution of Ricardo Barroso Campos
Universidad de Sevilla.
España
Let's consider the parallel one to BC for E, that will cut AC in V, that it is half point.
Let's plan the circumference circumscribed to the triangle AEV.
The perpendicular for V to YC is his peprndicular bisector that will cut to the
circumference limited to AEV in T.
The angle AET is straight. It is a question of seeing that D=T
T is suh that: TY=TC.
<TCV=<TYV
If we draw the circumference circumscribed to CTV:
Sea Z el punto de intersección de la circunscrita a YVT con EV
Es ZYV un triángulo semejante a XYZ y por tanto isósceles, por lo que YT es
perpendicular a ZV y a XC. Además es la mediatriz de XC.
Luego T es el circuncentro de YXC.
Así, el triángulo AED=AET , y al ser AD=AT un diámetro de la circunscrita al
triángulo AE, es <AED=90º.
Be Z the point of intersection of it circumscribed to YVT with EV
ZYV is a similar triangle to XYZ and therefore isosceles, for which YT is perpendicular
to ZV and to XC. Besides is the mediatriz of XC.
Then T is the circumcenter of YXC.
Thus, the triangle AED = AET, and to be AD = AT a diameter of it circumscribed to
triangle AE, is <AED=90º.