Art of Problem Solving

1983 AHSME Problems/Problem 28: Revision history

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20 February 2019

27 January 2019

9 July 2018

4 August 2017

  • curprev 08:0808:08, 4 August 2017 Mamis511 talk contribs 258 bytes +258 Created page with "Clearly since <math>[DBEF] = [ABE]</math> it follows that <math>[ADF] = [AFE]</math>. This implies that <math>AC \parallel DE</math> and so <math>\frac{AD}{DB} = \frac{CE}{EB}..."