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
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- 1. The area of △ AEF is equal to the area of trapezoidal EFBC
- 2. As shown in the figure, D is the midpoint of BC, the length of ad is three times that of AE, the length of EF is three times that of BF, the area of triangle AEF is 18 square centimeters, and the area of triangle ABC is 18 square centimeters What's the square centimeter?
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- 11. As shown in the figure, in the triangle ABC, the bisector of angle ABC and angle ACB intersects at point O, passes through the point and makes EF parallel BC, intersects AB at e, intersects AC at F, and △ AB As shown in the figure, in the triangle ABC, the bisector of the angle ABC and the angle ACB intersects at the point O, passes through the point for EF parallel BC, intersects AB at e, intersects AC at F, and △ ab.% d% a as shown in the figure, in the triangle ABC, the bisector of the angle ABC and the angle ACB intersects at the point O, passes through the point for EF parallel BC, intersects AB at e, intersects AC at F, and the perimeter of △ ABC is 24cm, BC = 10cm, calculate the perimeter of the triangle AEF
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