In parallelogram ABCD, the bisector of angle A intersects side BC at point F. Prove that triangle ABF is isosceles.

| March 3, 2021 | education

In parallelogram ABCD, the bisector of angle A intersects side BC at point F. Prove that triangle ABF is isosceles. Find the area AFCD if the angle BAD = 60 degrees. AB = 3, BC = 5.

The bisector AF divides the angle BAD in half, the angle BAF = DAF = 60/2 = 30.

Angle АFВ and DАF lying crosswise at the intersection of parallel АD and ВС secant АF, then angle BAF = BFA, and triangle ABF is isosceles, which was required to prove.

Angle ABC = (180 – 60) = 120.

In the triangle ABF, by the cosine theorem, we define the length AF.

AF ^ 2 = AB ^ 2 + BF ^ 2 – 2 * AB * BF * Cos120 = 9 + 9 – 2 * 3 * 3 * (-1/2) = 18 + 9 = 27.

AF = 3 * √3 cm.

Determine the area of ​​the triangle AFD.

Sаfд = AF * AD * Sin30 / 2 = (3 * √3 * 5 * 1/2) / 2 = 15 * √3 / 4 cm2.

Determine the area of ​​the triangle CDF.

Sсдf = CF * СD * Sin60 / 2 = (2 * 3 * √3 / 2) / 2 = 6 * √3 / 4 cm2.

Then Safcd = Safd + Scdf = (15 * √3 / 4) + (6 * √3 / 4) = 21 * √3 / 4 cm2.

Answer: The area of ​​the quadrangle АFСD is 21 * √3 / 4 cm2.



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