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

But try the problem first…

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

AMC-10A (2010) Problem 14

Pre College Mathematics

## Try with Hints

First hint

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……..

Second Hint

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

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……..

Final Step

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

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

## Other useful links

- https://www.youtube.com/watch?v=fvhDNPfri9w
- https://www.cheenta.com/problem-on-balls-isi-b-stat-objective-problem-128/

Google