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 (overline{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? (sqrt{(2a - 0)^2+(0 - 0)^2}=sqrt{(0 - 0)^2+(2b - 0)^2}=sqrt{(a - 0)^2+(b - 0)^2}) ((2a)^2=(2b)^2=(2c)^2) (2a = 2b = 2c) (sqrt{(0 - a)^2+(2b - b)^2}=sqrt{(2a - a)^2+(0 - b)^2}=sqrt{(a - 0)^2+(b - 0)^2})

Explanation:

Step1: Calculate distance DA

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

Step2: Calculate distance DB

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

Step3: Calculate distance DC

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

Step4: Compare distances

$DA = DB = DC$

Answer:

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