Question

Given ABC with A(0, 0), B(2, 3), and C(4,0), prove ABC is isosceles

Answer 1

Step-by-step explanation:

1. if to calculate the length of AB; AC and BC, then:

$|AB|=\sqrt{(2-0)^2+(3-0)^2} =\sqrt{13}.$

$|AC|=\sqrt{(4-0)^2+(0-0)^2} =4.$

$|BC|=\sqrt{(2-4)^2+(0-3)^2}= \sqrt{13}.$

2. if to compare the calculated lengths, then:

AB=BC. It means the given triangle is isosceles.