Given 4 points on a circle A, B, C, and P. Draw the orthogonal projections of P onto triangle ABC and call them $P_1, P_2,P_3$. Show that $P_1, P_2,P_3$ are collinear.
After drawing this out, I notice that the configuration looks exactly like the one for Menelaus' theorem. Unfortunately I am inexperienced with using it. Is this the right way to go about the problem or is there an easier different way?
On
It is well-known fact that these three points are collinear. They lie on so-called Simson line of point $P$ with respect to triangle $ABC$.
I think you can use a Menelaus' theorem here. Assume that $P_1 \in BC$, $P_2 \in CA$ and $P_3 \in AB$. Note that $\frac{BP_1}{P_1C} = \frac{\cot \angle PBC}{\cot \angle PCB}$ and so on. After multiplying this equality with two analogous ones, the right side will cancel out and you'll get $\frac{BP_1}{P_1C} \cdot \frac{CP_2}{P_2A} \cdot \frac{AP_3}{P_3B} = 1$.
Another way to prove it is noting that the following quadruples of points are concyclic: $(P,P_2,P_3,A)$, $(P, P_3, P_1, B)$ and $(P,P_1,P_2,C)$. Then it is just easy angle calculation.
The Simson theorem can be proved by angle chasing only.
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Since $BP_1 P P_2$ is a cyclic quadrilateral, $$\widehat{BP_2 P_1}=\widehat{BPP_1}=\widehat{ABP}-\frac{\pi}{2},$$
and since $CPP_2P_3$ is a cyclic quadrilateral, $$\widehat{CP_2 P_3}=\widehat{CPP_3}=\frac{\pi}{2}-\widehat{PCA},$$ so $\widehat{BP_2 P_1}=\widehat{CP_2 P_3}$ proves that $P_1,P_2,P_3$ are collinear.