S_{OAEC}= S_{BFDE}, points on sides of angle (Argentina 1998 OMA L1 p3)

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S_{OAEC}= S_{BFDE}, points on sides of angle (Argentina 1998 OMA L1 p3)

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#1 h May 10, 2020, 5:21 AM

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Points $A, B, C, D$ are considered on the sides of the vertex angle $O$ (see figure). Let $E$ be the point of intersection of segments $AD$ and $BC$ . The line parallel to $AB$ is drawn by $C$ and the line parallel to $CD$ is drawn by $A$ . Let $F$ be the point of intersection of the two drawn lines. Show that $S_{OAEC}= S_{BFDE}$

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This post has been edited 1 time. Last edited by parmenides51, May 11, 2020, 12:20 AM

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#2 h May 19, 2020, 1:08 PM

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Let $O(0,0),A(a,0),B(b,0),C(c,mc),D(d,md)$ .

Point $E(\frac{abc-abd+acd-bcd}{ac-bd},\frac{cdm(a-b)}{ac-bd})$ .
Point $F(a+c,mc)$ .

$S_{OAEC}=\frac{acm(ad-2bd+bc)}{2(ac-bd)}= S_{BFDE}$ .

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