In rectangle ABCD: AB = 6 cm, AD = 10 cm AK-bisector of angle A. Determine the middle line of the trapezoid AKCD

univerkov | July 23, 2021 | education

Since, by condition, AK is a bisector, then the angle ABK = DAK.

Angle AKB = DAK as cross-lying angles at the intersection of parallel straight lines ВС and АD secant AK.

Then the angle BAK = BKA, and therefore the triangle ABK is isosceles, AB = BK = 6 cm.

In a rectangle, the opposite sides are equal, then AB = CD = 6 cm, AD = BC = 10 cm, then the segment CK = AD – BK = 10 – 6 = 4 cm.

Let’s define the middle line of the trapezoid.

MP = (AD + CK) / 2 = (10 + 4) / 2 = 7 cm.

Answer: The middle line of the trapezoid is 7 cm.

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