Here are my submissions :

1) Let \( \Delta ABC \) have cevians \( \overline{AD} \) and \( \overline{CE} \), which meet at a point \( F \) inside the triangle. Prove that \( [\Delta ABC] \cdot [\Delta DEF] = [\Delta BDE] \cdot [\Delta AFC] \), where \( [ A ] \) denotes area of figure \( A \).

2) If a cevian \( \overline{AQ} \) of an equilateral triangle \( ABC \) is extended to meet the circumcircle at \( P \), prove that \( \dfrac{1}{\overline{PB}} + \dfrac{1}{\overline{PC}} = \dfrac{1}{\overline{PQ}} \).

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## Comments

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TopNewestQuestion 1: Extend segment BF to hit AC at T. Construct \(DE\) and call its intersection point with \(BE\) , \(M\). Notice that the desired relation is equivalent to \(\frac{[DEF]}{[BDE]} = \frac{[AFC]}{[ABC]} \implies \frac{FM}{MB} = \frac{TF}{TB} \). We will use mass points. Put weights of \(p, q, r\) on \(A, B, C\) respectively. This implies that \(T\) has a mass of \(p+r\) with implies that \[\frac{TF}{BF} = \frac{q}{p+r} \implies \frac{TF}{TB} = \frac{q}{p+q+r} \]

Notice that \(F\) has mass \(p+q+r\). This implies that \[\frac{MF}{MB} = \frac{q}{p+q+r} \] and we are done.

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Nice use of Barycentric Coordinates !

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Question 2: We know by Ptomely's, that \(AP = PB + PC\).

Then we just need to prove that \[\frac{1}{PB} + \frac{1}{PC} = \frac{AP}{PB \cdot PC} = \frac{1}{PQ} \implies \frac{BP}{AP} = \frac{PQ}{CP}\]

Observe that \(\angle ABP = \angle ACB = 60^{\circ} \) and \(\angle APC = \angle ABC = 60^{\circ} \) because they substend the same arc. Similarly, \(\angle CBP = \angle CAP \) substend the same arc. This implies that \(\triangle BPQ \sim APC \) and the desired ratio is given with these proportional lengths.

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@Alan Yan Do check problem no. 5 in RMO board-2 by Nihar Mahajan ! It's more interesting than these :). Link

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@Xuming Liang @Calvin Lin @Sualeh Asif Here are some simple problems ! Presently not very good at Geometry, but planning to improve.

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