Question

Suppose in a quadrilateral, the diagonals bisect the angles at the vertices. Prove that it is a rhombus.

Answer 1

Let the quadrilateral be ABCD with diagonals AC and BD intersecting at P.

Since they bisect each other and are perpendicular,

$△APB,△BPC,△CPD$ and $△DPA$ are right triangles.

They all are congruent by SAS.

$△APB$ is congruent to the $△CPB$ because they share common side BD,

Side AP and CP are congruent ( since P is the mid point of AC ) and the included angles are both right angles since they are all congruent, their third side ( the hypotenuse of each ) are congruent ( CPCTC ).

A rhombus is a quadrilateral in which all four sides are congruent, so ABCD is a rhombus.

Hence, the answer is proved.