The bisector of angle A of parallelogram ABCD intersects side BC at point K. Find the perimeter

| August 5, 2021 | education

The bisector of angle A of parallelogram ABCD intersects side BC at point K. Find the perimeter of the parallelogram if BK = 10, CK = 18.

Let a parallelogram ABCD be given, sides AB || CD, BC || AD, angles <A, <C – acute, <B, <D – obtuse. Let us draw the bisector of the angle <A, AK, it divides the side BC into segments ВK and KС. ВС = ВK + KС = 10 + 18 = 28 cm.
Since AK is a bisector, it means that by its property <BAK = <KAD. Angles <KAD = <BKA – as internal crosswise at parallel lines and secant.
Consider a triangle ABK, BK = 10 cm, <BAK = <BKA, which means an isosceles triangle with a base of AK and sides AB = BK = 10 cm.
Let’s define the perimeter of the parallelogram:
p = 2 * AB + 2 * BC = 2 * 10 + 2 * 28 = 68 cm.
Answer: the perimeter is 68 cm.



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