SOLUTION: In the triangle, AC=130cm. Angle FAC = Angle BED = Angle ABE. Also, Angle EBD = Angle DHF = Angle ACF. Find the area of triangle EBC> Diagram: https://imgur.com/a/JMl9yyM

Question 1149588: In the triangle, AC=130cm. Angle FAC = Angle BED = Angle ABE. Also, Angle EBD = Angle DHF = Angle ACF. Find the area of triangle EBC>
Diagram: https://imgur.com/a/JMl9yyM
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

(1) Triangle EBC is isosceles, because angles EBC (EBD) and ECB (ACF) are given to be congruent.

(2) Triangles CAF and BED are similar, because they have two pairs of congruent angles. In those two triangles, FC=50 and BD=40 are corresponding sides, so the ratio of similarity is 50:40 = 5:4.

(3) In those two triangles, EB and AC are also corresponding sides. Then, given AC=130, we can determine that EB=104.

Now you have the lengths of all three sides of triangle EBC, so you could find the area using Heron's formula.

Or you could drop an altitude of isosceles triangle EBC from E to side BC and, knowing that the foot of the altitude bisects side BC, find the altitude of EBC using the Pythagorean Theorem and then find the area of triangle EBC using the standard formula of one-half base times height.

Either way (or any other way) I leave the final calculations to you.

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