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Question Number 111477 by Aina Samuel Temidayo last updated on 03/Sep/20
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?
$$\mathrm{Triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{AB}=\mathrm{2}\centerdot\mathrm{AC}.\:\mathrm{Let} \\ $$$$\mathrm{D}\:\mathrm{and}\:\mathrm{E}\:\mathrm{be}\:\mathrm{on}\:\mathrm{AB}\:\mathrm{and}\:\mathrm{BC} \\ $$$$\mathrm{respectively}\:\mathrm{such}\:\mathrm{that}\:\angle\mathrm{BAE} \\ $$$$=\angle\mathrm{ACD}.\:\mathrm{Let}\:\mathrm{F}\:\mathrm{be}\:\mathrm{the}\:\mathrm{intersections}\:\mathrm{of} \\ $$$$\mathrm{segments}\:\mathrm{AE}\:\mathrm{and}\:\mathrm{CD},\:\mathrm{and}\:\mathrm{suppose} \\ $$$$\mathrm{that}\:\bigtriangleup\mathrm{CFE}\:\mathrm{is}\:\mathrm{equilateral}.\:\mathrm{What}\:\mathrm{is} \\ $$$$\angle\mathrm{ACB}? \\ $$
Answered by 1549442205PVT last updated on 04/Sep/20
Commented by Aina Samuel Temidayo last updated on 04/Sep/20
What about the solution?
$$\mathrm{What}\:\mathrm{about}\:\mathrm{the}\:\mathrm{solution}? \\ $$
Commented by 1549442205PVT last updated on 04/Sep/20
From the hypothesis triangle CEF is equilateral infer ECF^(�) =CFE^(�) =60°. ⇒AFD^(�) =CFE^(�) =60° (1) Put BAE^(�) =ACD^(�) =α we have ACB^(�) =60°+α(2).By the property of the exterior angle of the triangle , For ΔADF we have BDC^(�) =DFA^(�) +DAF^(�) =60°+α(3)(by (1)) For ΔACD we have BDC^(�) =BAC^(�) +ACD^(�) =BAC^(�) +α(4) From (3)(4)we infer BAC^(�) =60°.Hence denote by G the midpoint of AB.From the hypothesis AB=2AC infer AG=AC infer triangle ACG is equilateral.Hence CD=AG=BG which means triangle BCD is isosceles at D.This gives us BCD^(�) =CBD^(�) =(1/2)AGC^(�) =30° From that we get ABC^(�) =ACD^(�) +BCD^(�) =90°
$$\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{triangle}\:\mathrm{CEF}\:\mathrm{is} \\ $$$$\mathrm{equilateral}\:\mathrm{infer}\widehat {\:\mathrm{ECF}}=\widehat {\mathrm{CFE}}=\mathrm{60}°. \\ $$$$\Rightarrow\mathrm{A}\widehat {\mathrm{FD}}=\widehat {\mathrm{CFE}}=\mathrm{60}°\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Put}\:\widehat {\mathrm{BAE}}=\widehat {\mathrm{ACD}}=\alpha\:\mathrm{we}\:\mathrm{have} \\ $$$$\widehat {\mathrm{ACB}}=\mathrm{60}°+\alpha\left(\mathrm{2}\right).\mathrm{By}\:\mathrm{the}\:\mathrm{property}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{exterior}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:, \\ $$$$\mathrm{For}\:\Delta\mathrm{ADF}\:\mathrm{we}\:\mathrm{have} \\ $$$$\widehat {\mathrm{BDC}}=\widehat {\mathrm{DFA}}+\widehat {\mathrm{DAF}}=\mathrm{60}°+\alpha\left(\mathrm{3}\right)\left(\mathrm{by}\:\left(\mathrm{1}\right)\right) \\ $$$$\mathrm{For}\:\Delta\mathrm{ACD}\:\mathrm{we}\:\mathrm{have}\:\widehat {\mathrm{BDC}}=\widehat {\mathrm{BAC}}+\widehat {\mathrm{ACD}} \\ $$$$=\widehat {\mathrm{BAC}}+\alpha\left(\mathrm{4}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{3}\right)\left(\mathrm{4}\right)\mathrm{we}\:\mathrm{infer}\:\widehat {\mathrm{BAC}}=\mathrm{60}°.\mathrm{Hence} \\ $$$$\mathrm{denote}\:\mathrm{by}\:\mathrm{G}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{AB}.\mathrm{From} \\ $$$$\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{AB}=\mathrm{2AC}\:\mathrm{infer}\:\mathrm{AG}=\mathrm{AC} \\ $$$$\mathrm{infer}\:\mathrm{triangle}\:\mathrm{ACG}\:\mathrm{is}\:\mathrm{equilateral}.\mathrm{Hence} \\ $$$$\mathrm{CD}=\mathrm{AG}=\mathrm{BG}\:\mathrm{which}\:\mathrm{means}\:\mathrm{triangle} \\ $$$$\mathrm{BCD}\:\mathrm{is}\:\mathrm{isosceles}\:\mathrm{at}\:\mathrm{D}.\mathrm{This}\:\mathrm{gives}\:\mathrm{us} \\ $$$$\widehat {\mathrm{BCD}}=\widehat {\mathrm{CBD}}=\frac{\mathrm{1}}{\mathrm{2}}\widehat {\mathrm{AGC}}=\mathrm{30}° \\ $$$$\mathrm{From}\:\mathrm{that}\:\mathrm{we}\:\mathrm{get}\:\widehat {\mathrm{ABC}}=\widehat {\mathrm{ACD}}+\widehat {\mathrm{BCD}} \\ $$$$=\mathrm{90}° \\ $$
Commented by Aina Samuel Temidayo last updated on 04/Sep/20
Thanks.
$$\mathrm{Thanks}. \\ $$

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