# I need help finding the sides of the parallelogram

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 $\overline{DA}$.

• Could you please upload a picture of the question? I could not understand what $P$ and $\alpha$ is. – ArsenBerk Jun 18 '18 at 8:00
• $P$ 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 $E$. – Hanlon Jun 18 '18 at 8:07
• 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. – Hanlon Jun 18 '18 at 8:08
• Is $\alpha$ angle BAD or angle ADC? – Steve B Jun 18 '18 at 8:33
• 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. – Steve B Jun 18 '18 at 8:46

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 $\overline{AB}$ = $\overline{BE}$
So $\overline{AB}$ + $\overline{BE}$ + $\overline{EC}$ = 19.
2*$\overline{AB}$ + 3 = 19, so $\overline{AB}$ = 8
And $\overline{BC}$ = 8 + 3 = 11