In the parallelogram $ABCD$ the angle bisector line of the angle $\alpha$ cuts the segment $\overline{BC}$ in point $E$. If $\overline{CE}=3$ and $P=38$, find $\overline{AB}$,$\overline{BC}$,$\overline{CD}$ and $\overline{DA}$.
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$\begingroup$ Could you please upload a picture of the question? I could not understand what $P$ and $\alpha$ is. $\endgroup$ – ArsenBerk Jun 18 '18 at 8:00
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$\begingroup$ $P$ is the perimeter of the given parallelogram. $\alpha$ is simply the angle that a line bisects. That same line (if extended) cuts $\overline{BC}$ in point $E$. $\endgroup$ – Hanlon Jun 18 '18 at 8:07
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$\begingroup$ I can draw a picture in LaTeX but it will take too much time (and I don't have it much). I hope it's clear now what I'm asking. $\endgroup$ – Hanlon Jun 18 '18 at 8:08
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$\begingroup$ Is $\alpha$ angle BAD or angle ADC? $\endgroup$ – Steve B Jun 18 '18 at 8:33
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1$\begingroup$ Then the problem cannot be solved. If $\alpha$ is angle BAD, $\overline{AB}$ = 8 and $\overline{BC}$ is 11. If $\alpha$ is angle ADC, $\overline{AB}$ = 3 and $\overline{BC}$ is 16. $\endgroup$ – Steve B Jun 18 '18 at 8:46
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This shows the picture if $\alpha$ is $\angle$BAD. AE bisects $\angle$BAD so $\angle$BAE = $\angle$EAD. AD and BC are parallel, so $\angle$EAD = $\angle$AEB Which means $\angle$BAE = $\angle$AEB and AEB is isosceles. So $\overline{AB}$ = $\overline{BE}$
The perimeter is 38, making the semi-perimeter 19.
So $\overline{AB}$ + $\overline{BE}$ + $\overline{EC}$ = 19.
2*$\overline{AB}$ + 3 = 19, so $\overline{AB}$ = 8
And $\overline{BC}$ = 8 + 3 = 11
