Write a coordinate proof.
GIVEN: △ ABD is a right triangle, with the right angle at vertex A. Point C is the midpoint of hypotenuse BD.PROVE: Point C is the same distance from each vertex of △ ABD.

The correct answer is: the midpoint of the hypotenuse is equidistant from the vertices

Complete step by step solution:
Let ABD be the right triangle right angled at A. Here BD is the hypotenuse and
C is the midpoint of the hypotenuse so that BC= DC.
∠DAB = 90°

By the converse of the midpoint theorem, P is the midpoint of AD.
∠DPC = ∠DAB  = 90°(since PC ∥ AB)
Consider  ⃤  DPC and  ⃤  APC,
DP = AP (since P is the midpoint)
∠DPC = ∠APC  = 90°
CP = CP (common side)
∴ ⃤  DPC and  ⃤  APC are congruent by SAS congruence criterion.
⇒ DC = AC  (corresponding parts of congruent triangles)
Hence, DC = AC = CB
So, proved that the midpoint of the hypotenuse is equidistant from the vertices.