In the triangle ABC, the bisector of angle ACB intersects AB at e, EF parallels BC, intersects AC at F, and the bisector of the outer angle of angle ACB intersects G Try to judge the shape of the triangle EFC and explain your reason

In the triangle ABC, the bisector of angle ACB intersects AB at e, EF parallels BC, intersects AC at F, and the bisector of the outer angle of angle ACB intersects G Try to judge the shape of the triangle EFC and explain your reason

EF//BC, ∠FEC=∠ECB ∠FCE=∠ECB
∠FEC=∠FCE
Fe = FC, & nbsp; △ EFC is isosceles triangle
Here are some other points
∠GCE=∠GCA+∠ACE=∠DCA/2+∠ACB/2=(∠DCA+∠ACB)/2=90
∠G=90-∠FEC ∠FCG=90-∠FCE
∠G=∠FCG  FG=FC
Delta GCE is a right triangle and CF is the center line of hypotenuse