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Blaikie theorem:Let $O$ be any point in the plane of triangle $ABC$, and let any straight line g through $O$ meet $BC$ in $P, CA$ in $Q, AB$ in $R$, then, if points $P', Q', R'$ be taken on the line so that $ PO = OP', QO = OQ',RO = OR',$ then $AP', BQ', CR'$ are concurrent.

A generalization: Let $ABC$ be a triangle, let $L$ be a line on the plane. Let $L$ meets $BC, CA, AB$ at $A_0, B_0, C_0$. Let $X, Y, Z$ lie on $L$. Then $AX, BY, CZ$ are concurrent if only if:

$$\frac{\overline{XB_0}}{\overline{XC_0}}.\frac{\overline{YC_0}}{\overline{YA_0}}.\frac{\overline{ZA_0}}{\overline{ZB_0}}=1$$

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  • $\begingroup$ Dear Dr. @daijgrinberg $\endgroup$ Sep 20, 2015 at 7:23
  • $\begingroup$ Dear Mister @Blue $\endgroup$ Sep 20, 2015 at 7:25

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Consider two perspective triangles $ABC, XYZ$, let $A_0 = YZ\cap BC$, similarly define $B_0, C_0$. From Desargues theorem we get that $A_0, B_0, C_0$ are collinear and from Menelaus theorem that $\frac{XB_0}{XC_0}\ldots = 1$. Now in the degenerate case when $X, Y, Z$ are collinear we get proof of the main theorem. $\Box$

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