Medians $\overline{AX}$ and $\overline{BY}$ of $\triangle ABC$ are perpendicular at point $G$. Prove that $AB = CG$.
In your diagram, $\angle AGB$ should appear to be a right angle.
I've drawn the diagram, but I don't have anything in mind.
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Let $M -$ midpoint $AB$.
The median on the hypotenuse of a right triangle equals one-half the hypotenuse.
In triangle $AGB$ $\angle AGB=90^{\circ}$ then $GM=MA=MB=\frac12AB \Rightarrow AB=2GM$
The centroid divides each median into two segments, the segment joining the centroid to the vertex multiplied by two is equal to the length of the line segment joining the midpoint to the opposite side.
$CG:GM=2:1 \Rightarrow CG=2GM$.
Hence, $$CG=AB$$
Roman83
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How did you get $GM=\frac{1}{2}AB$? Also how did you get $CG:GM=2:1$? – Aug 12 '16 at 15:58
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http://math.stackexchange.com/questions/240819/whats-the-name-of-the-theoremmedian-of-right-triangle-hypotenuse-is-always-hal – Roman83 Aug 12 '16 at 16:01
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Thanks for the clarifications, Roman83! – Aug 12 '16 at 16:03
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http://www.vitutor.com/geometry/plane/orthocenter.html – Roman83 Aug 12 '16 at 16:04