Triangle's side
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$\triangle ABC$ is a equilateral triangle. $E$ and $D$ are points inside the triangle $\triangle ABC$ (not on the sides) and the line $DE$ is parallel to $BC$ .If $DE=1, BD=2,CE=2$ and $AD=AE=\sqrt {7}$.Find the length of the sides of the triangle?
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Re: Triangle's side
Let every side be a..
Draw a perpendicular line from vertex A to BC.It will intersect DE and BC in the point G an F respectively.Now we will find the area of the triangle ADE using the formula (b/4)*√(4a^2-b^2).It will come out 3√3/4.Then we will find the height of the triangle ADE.It will be 3√3/2.The length of AF will be √3/2a.From the problem we get that BCED is a trapezium.
So we can write the following equation:
area of triangle ADE+area of trapezium BCED+2*area of triangle AEC=√3a^2/4
I know it's a lengthy process.but i couldn't find other any.
Draw a perpendicular line from vertex A to BC.It will intersect DE and BC in the point G an F respectively.Now we will find the area of the triangle ADE using the formula (b/4)*√(4a^2-b^2).It will come out 3√3/4.Then we will find the height of the triangle ADE.It will be 3√3/2.The length of AF will be √3/2a.From the problem we get that BCED is a trapezium.
So we can write the following equation:
area of triangle ADE+area of trapezium BCED+2*area of triangle AEC=√3a^2/4
I know it's a lengthy process.but i couldn't find other any.