From the picture attached,
Coordinates of point A → (0, 0)
Coordinates of point M → \((\frac{a}{2}, \frac{b}{2} )\)
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 \((x_1,y_1)\) and \((x_2,y_2)\).
Distance = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
AM = \(\sqrt{(\frac{a}{2}-0 )^2+(\frac{b}{2}-0)^2}\)
= \(\sqrt{(\frac{a}{2})^2+(\frac{b}{2} )^2}\)
BM = \(\sqrt{(a-\frac{a}{2} )^2+(\frac{b}{2})^2}\)
= \(\sqrt{(\frac{a}{2})^2+(\frac{b}{2})^2}\)
CM = \(\sqrt{(0-\frac{a}{2} )^2+(b-\frac{b}{2})^2}\)
= \(\sqrt{(\frac{a}{2})^2+(\frac{b}{2})^2}\)
Therefore, AM = BM = CM = \(\sqrt{(\frac{a}{2})^2+(\frac{b}{2})^2}\)
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