In an isosceles trapezoid ABCD with bases AD and BC, the angle BDA = 45 degrees

In an isosceles trapezoid ABCD with bases AD and BC, the angle BDA = 45 degrees and the angle BDC = 43 degrees. Find the degree measure of the angle ABD.

In a trapezoid, the angle CBD is equal to the angle BDA, since these are cross-lying angles with parallel straight lines BC and AD and secant BD. That is, CBD = BDA = 45 degrees.

In a trapezoid, the angle BAD is equal to the angle CDA – these are the angles at the base of an isosceles trapezoid.

Find the angles at the base AD
The CDA angle is equal to the sum of the BDA and BDC angles.

angle CDA = 43 + 45 = 88 degrees.

The second angle at the base of the BAD is also 88 degrees.

The sum of the angle BAD and ABC is 180 degrees (properties of an isosceles trapezoid).

Find the unknown angle ABC.

BAD + ABC = 180

ABC = 180 – BAD

ABC = 180 – 88 = 92 degrees.

Angle ABC is equal to the sum of angles ABD and CBD. It turns out equality with one unknown.

ABC = ABD + СВD

92 = ABD + 45

ABD = 92 – 45 = 47

Answer: The ABD angle is 47 degrees.