Find .
Note : The figure is not drawn to scale.
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Draw 2 lines, AE and FC. AE//FC because AF = FD, and EC = CD. (Mid-point theorem)
Also, if FC = k, AE = 2k.
Since AE//FC,
∠AEF = ∠EFC (alt. ∠s, AE//FC)
∠EAC = ∠ACF (alt. ∠s, AE//FC)
∠ABE = ∠CBF (vert. opp. ∠s)
(Choose any 2 of the above)
∴ △BCF~△BAE (AA or AAA)
Since △BCF~△BAE (proved), and AE = 2 x FC,
AB = 2 x BC
∴ AB : BC = 2 : 1