In parallelogram ABCD, the bisector of angle BAD is drawn, which intersects side BC at point m.
| January 16, 2021 | education
In parallelogram ABCD, the bisector of angle BAD is drawn, which intersects side BC at point m. a) prove that triangle ABM is isosceles b) Find side AD if BM is 8cm and the parallelogram perimeter is 38cm.
Since AM is the bisector of the angle, then the angle BAM = DAM.
The angle DAM = BMA as the cross-lying angles at the intersection are parallel to the lines AD and BC of the secant AM, then the angle BAM = BMA, and therefore the triangle ABM is isosceles with the base AM, which was required to be proved.
Since the triangle ABM is isosceles, then AB = BM = 8 cm, then CD = 8 cm.
Ravsd = 2 * (AD + AB) = 38 cm.
BP + AB = 38/2 = 19 cm.
AD = 19 – AB = 19 – 8 = 11 cm.
Answer: The length of the AD side is 11 cm.
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