Triangle ABC has AB=2∙AC. Let D and E be on AB and BC respectively such that ∠BAE =∠ACD. Let F be the intersections of segments AE and CD, and suppose that △CFE is equilateral. What is ∠ACB?


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Question Number 111477 by Aina Samuel Temidayo last updated on 03/Sep/20

$Triangle ABC has AB = 2 \cdot AC . Let$ $D and E be on AB and BC$ $respectively such that ∠ BAE$ $= ∠ ACD . Let F be the intersections of$ $segments AE and CD, and suppose$ $that △ CFE is equilateral . What is$ $∠ ACB ?$
	Answered by 1549442205PVT last updated on 04/Sep/20

	Commented by Aina Samuel Temidayo last updated on 04/Sep/20

	$What about the solution ?$
	Commented by 1549442205PVT last updated on 04/Sep/20

	$From the hypothesis triangle CEF is$ $equilateral infer \hat{ECF} = \hat{CFE} = 60 ° .$ $\Rightarrow A \hat{FD} = \hat{CFE} = 60 ° (1)$ $Put \hat{BAE} = \hat{ACD} = α we have$ $\hat{ACB} = 60 ° + α (2) . By the property of the$ $exterior angle of the triangle,$ $For Δ ADF we have$ $\hat{BDC} = \hat{DFA} + \hat{DAF} = 60 ° + α (3) (by (1))$ $For Δ ACD we have \hat{BDC} = \hat{BAC} + \hat{ACD}$ $= \hat{BAC} + α (4)$ $From (3) (4) we infer \hat{BAC} = 60 ° . Hence$ $denote by G the midpoint of AB . From$ $the hypothesis AB = 2 AC infer AG = AC$ $infer triangle ACG is equilateral . Hence$ $CD = AG = BG which means triangle$ $BCD is isosceles at D . This gives us$ $\hat{BCD} = \hat{CBD} = \frac{1}{2} \hat{AGC} = 30 °$ $From that we get \hat{ABC} = \hat{ACD} + \hat{BCD}$ $= 90 °$
	Commented by Aina Samuel Temidayo last updated on 04/Sep/20

	$Thanks .$
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