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-

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 .