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Quadrilateral diagram

In the image above, coordinates of points $A,\, B,\, C,\, D$ and $E$, are known. Also line $AB$ is parallel to $EX_1$ and so are lines $AC$ and $EX_2$. How can I find coordinates of points $X_1$ and $X_2$.

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  • $\begingroup$ Exploit the obvious similarities. $\endgroup$ – Jack D'Aurizio Jun 30 '17 at 21:14
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Let $r = \frac {DE}{DA}= \frac {DX_2}{DC}= \frac {DX_1}{DB}$.

Then $X_1$ and $X_2$ are $r$ of the distance between $D$ and $B$ and between $D$ and $C$.

So $X_1= (D_x + r(B_x - D_x), D_y + r(B_y- D_y))$ and $X_2 =(D_x + r(C_x - D_x), D_y + r(C_y- D_y))$.

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  • $\begingroup$ Hey, can you please explain/elaborate your method? $\endgroup$ – Angad Singh Jun 30 '17 at 21:19
  • $\begingroup$ As Jack D'Aurizio said "Exploit the obvious similarities". Obviously $r = \frac {DX_1}{DB} = \frac {DE}{DA}= \frac {DX_2}{DC}$. So my result, once I fix the typos, should be obvious. $\endgroup$ – fleablood Jun 30 '17 at 21:23
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Hint: You can find equations of the lines $AB$ and $BD$. Then you find the line parallel to $AB$ and passes through $E$. Finally, you find intersection of the lines $BD$ and $EX_1$.

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Working in homogeneous coordinates, $$X_1\sim (B\times D)\times((A-B)\times E) \\ X_2\sim (C\times D)\times((A-C)\times E).$$ This is basically Farrukh Ataev’s answer computed directly.

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