In rectangle ABCD, AB = 6, AD = 10, AK is the bisector of angle A. Find the midline of the resulting trapezoid.

univerkov | February 27, 2021 | education

Since AK is the bisector of the angle BAD, the angle BAK = DAK.

The angle DАК is equal to the angle ВКА as criss-crossing angles at the intersection of parallel straight lines ВС and АD secant АС. Then the ВKA angle is equal to the ВAK angle, and therefore the AВK triangle is isosceles and AB = ВK = 6 cm.

Since ABCD is a rectangle, then AD = BC = 10 cm, then СK = BC – ВK = 10 – 6 = 4 cm.

Determine the length of the midline of the MР trapezoid.

MР = (AD + KС) / 2 = (10 + 4) / 2 = 14/2 = 7 cm.

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

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