Problem
GEO-B2-M09-P003 Midpoint of the Hypotenuse
#3
★☆☆☆☆ Level 1 of 5
In triangle \(A(0,0)\), \(B(m,0)\), \(C(0,n)\), where \(m,n>0\), point \(M\) is the midpoint of \(BC\). Prove that \(MA=MB=MC\).
Find the coordinates of \(M\) and compare squared distances.
We have \(M\left(\frac m2,\frac n2\right)\). Then \(MA^2=\frac{m^2+n^2}{4}\), \(MB^2=\left(m-\frac m2\right)^2+\left(0-\frac n2\right)^2=\frac{m^2+n^2}{4}\), and similarly \(MC^2=\frac{m^2+n^2}{4}\). Thus all three distances are equal.
It is important to compare squared lengths without taking square roots.