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Prove that $PA^2\sin A + PB^2 \sin B + PC^2 \sin C$ takes the same value for all points P on the incircle of the triangle $\triangle ABC$.

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The previous part asks that

If I is the incentre of the $\triangle ABC$ and $\alpha$, $\beta$, $\gamma$ are respectively the angles $BIC$, $CIA$ and $AIB$, prove that $$\frac{a \cdot IA}{\sin \alpha} = \frac{b \cdot IB}{\sin \beta} = \frac{c \cdot IC}{\sin \gamma}$$

And I have solve it with trigonometry, but I am not sure how to use it to prove the first statement. The general problem can be found here but I wonder if there is a simpler solution here since it only asks for points on the incircle. An idea is to use the fact that $$PA^2 \sin A = PA \cdot NL$$ etc. But I am not sure how to proceed.

Y.T.
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  • Presumably $LMN$ is the pedal triangle of $P$? – user10354138 May 11 '21 at 12:22
  • @user10354138 Yes – Y.T. May 11 '21 at 12:25
  • Do you accept a proof using vectors as an answer to your question? – YNK May 17 '21 at 18:47
  • @YNK sure if it is any simpler compared to the linked answer. – Y.T. May 18 '21 at 04:48
  • One of the answers in the linked reference, I am sure you know which one, proves that the statement in question is a minimum when $P$ coincides with the incenter. But it also contains everything one needs to prove the statement in your question. I thought of showing you that by elaborating (expanding and simplifying) certain parts of that answer. Now, I know that that is not what you need. So, I am not posting anything. – YNK May 18 '21 at 09:23

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