i) Let's extend the line CA in the direction of A so that AB=AF. In this case, m(ABF)=m(BFC)=20 and BF=BC.(Triangle BFC,20-20-140) ii) Let's choose a point K outside the triangle and an E point on FC on the AC closed line so that FCnBK=E and m(ABE)=40. In this case, triangle BFK is equilateral triangle (triangle FKE, 40-60-80) is obtained. Also, AE=BE. iii) When K and D are combined, since EKD is an isosceles triangle (40-40-140)m(FDK)=40, FK=KD. in this case, BK=KD and the triangle BKD is 40-70-70 triangle. So x+40=70 from here x=30 degrees
i) Let's extend the line CA in the direction of A so that AB=AF. In this case, m(ABF)=m(BFC)=20 and BF=BC.(Triangle BFC,20-20-140)
ReplyDeleteii) Let's choose a point K outside the triangle and an E point on FC on the AC closed line so that FCnBK=E and m(ABE)=40. In this case, triangle BFK is equilateral triangle (triangle FKE, 40-60-80) is obtained. Also, AE=BE.
iii) When K and D are combined, since EKD is an isosceles triangle (40-40-140)m(FDK)=40, FK=KD. in this case, BK=KD and the triangle BKD is 40-70-70 triangle. So x+40=70 from here x=30 degrees