In a parallelogram ABCD, the bisectors of angles A and B meet at point K. Prove that angle AKB is a right angle.

In a parallelogram ABCD, consecutive angles are supplementary. This means that the sum of adjacent angles is $180^\circ$.
Therefore, $\angle DAB + \angle ABC = 180^\circ$.

AK bisects $\angle DAB$, which means it divides the angle into two equal parts.
So, $\angle KAB = \frac{1}{2} \angle DAB$.
BK bisects $\angle ABC$, which means it divides the angle into two equal parts.
So, $\angle KBA = \frac{1}{2} \angle ABC$.

Divide the equation from Step 1 by 2:

Substitute the expressions for $\angle KAB$ and $\angle KBA$ from Step 2 into this equation:

The sum of angles in any triangle is $180^\circ$. For $\triangle AKB$:

Substitute the result from Step 3 into the equation from Step 4:

Subtract $90^\circ$ from both sides:

Since $\angle AKB = 90^\circ$, it means that line segment AK is perpendicular to line segment BK.