The bisector of angle A of parallelogram ABCD intersects side BC at point K. Find the perimeter
| August 15, 2021 | education
The bisector of angle A of parallelogram ABCD intersects side BC at point K. Find the perimeter of the parallelogram if BK = 3, CK = 19.
Given: ABCD – parallelogram, AK – bisector, ВK = 3, СK = 19.
Find: PAВСD -?
Solution:
Because AK is a bisector, then ∠ ВAK = ∠ КАD. Consider a triangle ABK. ∠ КАD = ∠ BKA (intersecting, with parallel lines ВС and АD and secant АD). Therefore, ∠ BKA = ∠ BAK, triangle ABK is isosceles, AB = BK = 3.
By the property of the parallelogram AB = CD = 3. BC = BK + KC = 3 + 19 = 22. BC = AD = 22.
P = 2 * AB + 2 * AD = 2 * 3 + 2 * 22 = 6 + 44 = 50.
Answer: 50.
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