Chapter 7.1, Problem 19PSC

To determine

To prove : The midpoint of a the hypotenuse of a right triangle is equidistant from all three vertices.

Explanation of Solution

Given information : The following information has been given

ABC is a triangle with AB as hypotenuse

D is the midpoint of AB

Formula used : Diameter of a circle makes a ${90}^0$ triangle with any point on the circle.

Proof : We know that D is the midpoint of AB, hence, we can say that

  $\overrightarrow{DA}\cong \overrightarrow{DB}$

Now, let’s imagine a circle with center at D and AB as diameter

We know that diameter extends a ${90}^0$ at any point on the circle if it makes a triangle. Let’s assume C is such a point on the circle we diameter extends a ${90}^0$ angle.

Thus, we know that as C is on the circle,

  $\overrightarrow{DA}\cong \overrightarrow{DB}\cong \overrightarrow{DC}$

Hence, proved that the midpoint of a the hypotenuse of a right triangle is equidistant from all three vertices