MHB Proving isosceles using centroid and medians

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The discussion focuses on proving the isosceles triangle property using centroids and medians. By letting sides AB, AC, and BC be represented as c, b, and a respectively, the centroid properties are applied alongside Stewart's Theorem to derive an equation involving the triangle's sides. The resulting equation simplifies to $(b-c)^2 \left[ (b-c)^2-a^2 \right] =0$. The analysis reveals that the bracket fails the triangle inequality, leading to the conclusion that b must equal c, indicating that AB equals AC. The author expresses difficulty in finding a synthetic proof for this result.
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I can definitely do this in the opposite direction, but
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Letting $AB=c$, $AC=b$ and noting that the usual property of the centroid tells us that $GC = \frac{2}{3}$ of the median from $C$ and $GB = \frac{2}{3}$ of the median from $B$ we can use Stewart's Theorem to write everything in terms of the three sides $a,b,c$. It is an involved equation requiring two squarings to get rid of all radicals. I used Wolfram Alpha to double check my computations. After simplifying everything I end up with

$(b-c)^2 \left[ (b-c)^2-a^2 \right] =0$.

The bracket fails the triangle inequality which only leaves $b=c$ or $AB=AC$. I can't see any synthetic proof.
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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