Question

△abc is a right triangle with vertices at a(0,0), b(0,2b) and c(2a,0). point d(a,b) is the midpoint of bc. which of the following statements could be used to show that the midpoint of the hypotenuse of any right triangle is equidistant from the vertices of the triangle?

Explanation:

Step1: Recall distance formula

$d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$

Step2: Calculate $DA$

$DA=\sqrt{(a - 0)^2+(b - 0)^2}=\sqrt{a^{2}+b^{2}}$

Step3: Calculate $DB$

$DB=\sqrt{(a - 0)^2+(b - 2b)^2}=\sqrt{a^{2}+b^{2}}$

Step4: Calculate $DC$

$DC=\sqrt{(a - 2a)^2+(b - 0)^2}=\sqrt{a^{2}+b^{2}}$

Step5: Conclude

Since $DA = DB = DC$, the mid - point of the hypotenuse is equidistant from the vertices.

Answer:

We can use the distance - formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$ to show that $DA=DB = DC$.

1. Calculate the distance between $D(a,b)$ and $A(0,0)$:

• By the distance - formula, $DA=\sqrt{(a - 0)^2+(b - 0)^2}=\sqrt{a^{2}+b^{2}}$.

1. Calculate the distance between $D(a,b)$ and $B(0,2b)$:

• $DB=\sqrt{(a - 0)^2+(b - 2b)^2}=\sqrt{a^{2}+(-b)^{2}}=\sqrt{a^{2}+b^{2}}$.

1. Calculate the distance between $D(a,b)$ and $C(2a,0)$:

• $DC=\sqrt{(a - 2a)^2+(b - 0)^2}=\sqrt{(-a)^{2}+b^{2}}=\sqrt{a^{2}+b^{2}}$.

Since $DA = DB=DC=\sqrt{a^{2}+b^{2}}$, the mid - point $D$ of the hypotenuse $BC$ is equidistant from the vertices $A$, $B$, and $C$ of the right - triangle $\triangle ABC$.