From the picture attached,
Coordinates of point A → (0, 0)
Coordinates of point M → [tex](\frac{a}{2}, \frac{b}{2} )[/tex]
Coordinates of point A → (0, b)
Coordinates of point A → (a, 0)
Since, we have to prove that midpoint (M) of the hypotenuse BC is equidistant from all three vertices A, B and C,
AM = BM = CM
Use the expression for the distance between two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex].
Distance = [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
AM = [tex]\sqrt{(\frac{a}{2}-0 )^2+(\frac{b}{2}-0)^2}[/tex]
= [tex]\sqrt{(\frac{a}{2})^2+(\frac{b}{2} )^2}[/tex]
BM = [tex]\sqrt{(a-\frac{a}{2} )^2+(\frac{b}{2})^2}[/tex]
= [tex]\sqrt{(\frac{a}{2})^2+(\frac{b}{2})^2}[/tex]
CM = [tex]\sqrt{(0-\frac{a}{2} )^2+(b-\frac{b}{2})^2}[/tex]
= [tex]\sqrt{(\frac{a}{2})^2+(\frac{b}{2})^2}[/tex]
Therefore, AM = BM = CM = [tex]\sqrt{(\frac{a}{2})^2+(\frac{b}{2})^2}[/tex]
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