. On segment there is a point such that halves the angle . If is the height of the of the triangle and . Then
In an isosceles triangle angle
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Angle C=108° we can calculate the other ones: (180-108)/2=A So angle A=36°. Now the angle CAE=18° and from there angle CAE=54°=ECA so triangle CHE is isosceles as well. Now Let FD be the middle line of the triangle ABE. Beacuse its the middle line it is parallel to BC and angle ADF=ABE=36° now the angle EFD=54°(ADF+FAD) Angle FHD=CHE =>CEH=HFD so the triangle FHD is isosceles as well. from there we can see that FH+HE=CH+HD or CD=EF. since AE=2EF so is AE=2CD. From there we get AE=2*9=18