Consider a parallelogram with a right angle and diagonals that bisect the angles.
In the figure, ABCD is a parallelogram with ∠A = 90°.
Since the opposite angles of a parallelogram are equal,
∠A = ∠C = 90° and ∠B = ∠D
Also, since adjacent angles of a parallelogram are supplementary,
∠A + ∠B = 180°
But, since ∠A = 90°, ∠B = 90° and ∠D = 90°
Therefore, ∠A = ∠B = ∠C = ∠D.
Now, it is given that the diagonals bisect the angles.
Therefore, ∠OAB = ∠OBA = ∠OBC = ∠OCB = 45°
Consider, triangles OBA and OBC.
∠OBA = ∠OBC = 45°
and OB = OB (common)
Therefore, Δ OBA ≅ Δ OBC (SAS Rule)
By corresponding parts of congruent triangles,
AB = BC
Note that in a parallelogram, the opposite angles are congruent.
Therefore, AB = BC = CD = DA.
Hence, in the parallelogram ABCD, we have,
AB = BC = CD = DA and ∠A = ∠B = ∠C = ∠D = 90°
Hence, ABCD is a square.
Answer:
A Parallelogram with a right angle and diagonals that bisect the angles
Step-by-step explanation:
I just took the test
