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May 7, 2020

Problem on Equilateral Triangle | AMC-10A, 2010 | Problem 14

Try this beautiful Geometry Problem on Equilateral Triangle from AMC-10A, 2010.

Equilateral Triangle - AMC-10A, 2010- Problem 14


Triangle $ABC$ has $AB=2 \cdot AC$. Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC}$, respectively, such that $\angle BAE = \angle ACD$. Let $F$ be the intersection of segments $AE$ and $CD$, and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB$?

  • \(60^{\circ}\)
  • \(70^{\circ}\)
  • \(90^{\circ}\)
  • \(75^{\circ}\)
  • \(1200^{\circ}\)

Key Concepts


Geometry

Triangle

Angle

Check the Answer


Answer: \(90^{\circ}\)

AMC-10A (2010) Problem 14

Pre College Mathematics

Try with Hints


Problem on equilateral triangle

We have to find out the \(\angle ACB\).Given that \(\angle CEF\) is a equilateral triangle and also given that $\angle BAE = \angle ACD$.so using the help of this two conditions ,we can find out all possible values of angles.........

can you finish the problem........

triangle figure

\(\angle BAE=\angle ACD=X\)

Let,

\(\angle BAE=\angle ACD=X\)

\(\angle BCD=\angle AEC=60^{\circ}\)

\(\angle EAC +\angle FCA+ \angle ECF+\angle AEC=\angle EAC +x+60^{\circ}+60^{\circ}=180^{\circ}\)

\(\angle EAC=60^{\circ}-x\)

\(\angle BAC =\angle EAC +\angle BAE =60^{\circ} -x+x=60^{\circ}\)

can you finish the problem........

Since \(\frac{AC}{AB}=\frac{1}{2} \angle BCA\)=\(90^{\circ}\)

Therefore value of \(\angle BCA=90^{\circ}\)

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