<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) 
$
...??

 [/INST] Here's a proof for the general case where $\small PC$ is not necessarily perpendicular to $\small AB$:
Given that we have already proven $$\mathsf{4.(sum\ of\ squares\ of\ medians)=3.(sum\ of\ squares\ of\ sides)}$$ using Appollonius theorem, continue from there.
So let's calculate $\small AP^2+AC^2+PC^2\tag{*}$
Drop the perp from $\small A$ to $\small PC$ and the foot $\small I$ must lie on the small circle (angle of half circle=$\small 90^\circ$).

Now using power of point,
$\small \begin{align}PI\cdot PB&=3\cdot 5\\ PI\cdot(PI+IB)&=15\\PI^2+PI\cdot IB&=15\end{align}$
See $\small PI=BC$.
Also note that $\small PI\cdot IC=PI\cdot(PI+IB)$.
From Pythagoras' theorem $\small AP^2=AI^2+PI^2$, $\small AC^2=AI^2+IC^2$ and $\small AB^2=AI^2+IB^2$.
Now substitute these in (*).
$\small \begin{align}&=AP^2+AC^2+PI^2+IC^2+2\cdot PI\cdot IC\\ &=2AI^2+2PI^2+2IC^2+30\\&=2(AB^2-IB^2+PI^2+IC^2+15)\\&=2(2^2+PI^2+(IC^2-IB^2)+15)\\&=2(4+PI^2+(IC-IB)(IC+IB)+15)\\&=2(4+PI^2+PI(PI+2IB)+15)\\&=2(4+2(PI^2+PI\cdot IB)+15)\\&=2(4+2\cdot15+15)\\&=98\end{align}$
Now using the known identity, sum of squares of medians
$=98\cdot\dfrac34=73.5$
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