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- TL;DR Summary
- Finding precise location of triangle in a right triangle with the give conditions

I would like to find the triangle ##\bigtriangleup DEF## in the plot below that is inside the right triangle ##\bigtriangleup ABC## given ##\overline{AB}=3, \overline{AC}=4## with ##\overline{BD}=\overline{DE},\overline{AE}=\overline{EF}, \overline{FC}=2\overline{DF}##. However, I'm finding it difficult to actually find precisely in the larger triangle where it is (diagram below is only approx.) . Surely if I let ##D=(x_3,y_3),F=(x_2,y_2), E=(x_1,y_1)## as per the diagram and solve the following six simultaneous equation for the coordinates of D,E, and F inside the larger triangle should do it but when I attempt to numerically solve the equations (in Mathematica), I obtain the empty set. However, the triangle does exist. I was wondering if someone could look at my work and confirm that I have a valid set of equations to find these points? For example, the first equation is obtained by noting:

## x_1^2+y_1^2=z^2## and ##(x_2-x_1)^2+(y_2-y_1)^2=z^2## and so forth.

Thanks guys.

$$\begin{array}{l}

x_1^2+y_1^2=(x_2-x_1)^2+(y_2-y_1)^2 \\

x_3^2+(3-y_3)^2=(x_3-x_1)^2+(y_3-y_1)^2 \\

4[(x_3-x_2)^2+(y_3-y_2)^2]=(4-x_2)^2+y_2^2 \\

x_2^2+y_2^2=4(x_1^2+y_1^2)\\

4[x_3^2+(3-y_3)^2]=x_1^2+(3-y_1)^2\\

9[(x_3-y_2)^2+(y_3-y_2)^2]=(4-x_3)^2+y_3^2

\end{array}

$$

## x_1^2+y_1^2=z^2## and ##(x_2-x_1)^2+(y_2-y_1)^2=z^2## and so forth.

Thanks guys.

$$\begin{array}{l}

x_1^2+y_1^2=(x_2-x_1)^2+(y_2-y_1)^2 \\

x_3^2+(3-y_3)^2=(x_3-x_1)^2+(y_3-y_1)^2 \\

4[(x_3-x_2)^2+(y_3-y_2)^2]=(4-x_2)^2+y_2^2 \\

x_2^2+y_2^2=4(x_1^2+y_1^2)\\

4[x_3^2+(3-y_3)^2]=x_1^2+(3-y_1)^2\\

9[(x_3-y_2)^2+(y_3-y_2)^2]=(4-x_3)^2+y_3^2

\end{array}

$$

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