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

Try this beautiful Geometry Problem on Equilateral Triangle from AMC-10A, 2010.You may use sequential hints to solve the problem.

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

Answer: $90^{\circ}$

AMC-10A (2010) Problem 14

Pre College Mathematics

## Try with Hints

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……..

$\angle BAE=\angle ACD=X$

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 $\angle BCA=90^{\circ}$