The problem states that the diagonals of a quadrilateral ABCD bisect each other. This is a key property that defines a parallelogram.
A quadrilateral whose diagonals bisect each other is always a parallelogram. Therefore, ABCD is a parallelogram.
In any parallelogram, consecutive (adjacent) angles are supplementary. This means their sum is equal to $180^{\circ}$.
So, for parallelogram ABCD:
We are given that $\angle A = 45^{\circ}$. Using the property that adjacent angles are supplementary:
$ \angle A + \angle B = 180^{\circ} $
Substitute the value of $\angle A$:
$ 45^{\circ} + \angle B = 180^{\circ} $
To find $\angle B$, subtract $45^{\circ}$ from both sides:
$ \angle B = 180^{\circ} - 45^{\circ} $
$ \angle B = 135^{\circ} $
Therefore, the measure of angle B is $135^{\circ}$.
What is the value of AC 2– BD 2
What is the point of intersection of the diagonals?
What is the area of the parallelogram?
ABCD is a cyclic quadrilateral. Diagonals BD and AC intersect each other at E. If ∠BEC = 138° and ∠ECD = 35°, then what is the measure of ∠BAC?
A circle is inscribed in a quadrilateral ABCD, touching sides AB, BC CD and DA at P, Q, R and S, respectively. If AS = 6 cm, BC = 12 cm, and CR = 5 cm, then the length of AB (in cm) is: