<s>[INST] Find the sum of the squares of the medians of triangle APC below For reference: The diameter of a circle measures 8m. On this diameter are located the points A and B equidistant 1 m from the center; through B draw any chord PC , determine the sum of the squares of the medians of triangle APC.(Answer: $73,5$)
My progress:
 sum of the squares of the median
$4(PE^2+CD^2+AB^2) = 3(AC^2+AP^2+CP^2)\therefore\\
PE^2+CD^2+AB^2=\frac{3}{4}(AC^2+AP^2+CP^2)\\
AC = AP \implies PE^2+CD^2+AB^2=\frac{3}{4}(2AC^2+CP^2)\\
CB=\frac{CP}{2}  \\
\triangle ABC: 4+(\frac{CP}{2})^2=AC^2\implies \frac{16+CP^2}{4} =AC^2\\
\therefore PE^2+CD^2+AB^2 = \frac{3}{8}(16+3CP^2) 
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 [/INST] If $O$ is the center of the circle, $OP$ is median of $\triangle APB$. Radius of the circle is $4$.
So, $4 OP^2 = 64 = 2 AP^2 + 2 PB^2 - AB^2$
$ \implies AP^2 + PB^2 = 34$
Similarly, $AC^2 + BC^2 = 34$
Now by intersecting chords theorem, $PB \cdot BC = 5 \cdot 3 = 15$
So, $AP^2 + AC^2 + PB^2 + BC^2 + 2 PB \cdot BC = 98$
$\implies AP^2 + AC^2 + CP^2 = 98$
Given you know the sum of squares of the sides of the triangle, you can find the sum of squares of the medians.
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