Is it possible to draw trapezoid by compass and straightedge if 4 sides given? I tried to construct trapezoid with lengths of 4 sides given. Without success. Then searched in internet. I believe it is not possible, but I am not sure.
The only way is to calculate the high of the trapezoid (distance of two paralell sides).
Is my assuption correct or there is a way how to construct?
 A: It is possible.
Let $a,b,c,d$ be the lengths of the trapezoid where $a$ and $c$ are the sides to be paralel. Without loss of generality let $a > c$.
Construct the segment $AB$ with length $a$.
Draw a circle centered in $A$ with radius $d$ and another one in $B$ with radius $b$. Call this circles $\pi_A$ and $\pi_B$.
Find the point $P$ in the segment $AB$ such that the distance $AP$ is $c$.
Find the point $Q$ in the segment $AB$ such that the distance $QB$ is $c$.
Draw a circle centered in $P$ with radius $d$. Call that circle $\pi_P$.
Draw a circle centered in $Q$ with radius $b$. Call that circle $\pi_Q$.
Now $\pi_B$ and $\pi_P$ intersect in two points. Call $C$ one of them. Now $\pi_A$ and $\pi_Q$ intersect in two points, one of wich is in the same semiplane as $C$ with respect to the segment $AB$. Call that point $D$.
Then $AB$ measures $a$ by construction, $BC$ measures $b$ by construction and $DA$ measures $d$ by construction.
It's not too hard to prove that $CD$ measures $c$ and is paralel to $AB$.
A: Let: 


*

*$\Gamma_A$ a circle with centre in $A$ and radius $AD$;

*$\Gamma_B$ a circle with centre in $B$ and radius $BC$;

*$v$ a vector parallel to $AB$ with length $CD$;



To find $C$, we may just intersect $\Gamma_A+v$ with $\Gamma_B$.
A: Yes. If $a,b,c,d$ are given and we want $a\|c$ and have $a>c$, then construct the triangle $APD$ with sides $AP=a-c$, $PD=b$, $DA=d$. Extend $AP$ beyond $P$ until $B$ so that $AB=a$. The parallel to $PD$ through $B$ and the parallel to $AP$ through $D$ intersect in $C$
A: We take an algebraic approach.
The height is constructible, given the sides and the information about which pair of sides is parallel. If $a$ and $b$ are the lengths of the parallel sides, assumed unequal, then the height is given by (please see Wikipedia)
$$h=\frac{\sqrt{(-a+b+c+d)(a-b+c+d)(a-b+c-d)(a-b-c+d}}{2|b-a|}.$$
Since only arithmetic operations and square root are involved, $h$ is straightedge and compass constructible given $a,b,c,d$. Now as you observed we have enough information to construct the trapezoid.
A: In ABCD, we assume that (1) $AD // BC$; and (2) WLOG, $BC \ge AD$.

Step-1 On the line segment BC [given], locate point E such that $BE = BC – AD$ (a simple and possible construction, hence skipped.)
Step-2 Construct the $\triangle ABE$ where $AE = DC$ [given]
Step-3 Construct the parallelogram AECD.
A: A person asking this question VERY likely is taking geometry and has not yet had algebra 2 or higher math. 
Draw a trapezoid given 4 sides. 
Lets call the given bottom  baseline B and the given top baseline T.
Lets call the given left side of the trapezoid L and the given right side of the trapezoid R. 
If it was not specified which side was left or right, you get to choose. 
First draw in the lower baseline as you want it. 
Likely this will be horizontal with plenty of room above it to draw in your trapezoid.
Make sure the length is of length B.  
Mark off on the lower baseline B  starting from the right hand side endpoint a distance equal to line T. Lets call this segment you marked off U. 
Lets call the distance you have left  D for distance. 
So you have broken line B into two line segments,  D to the left and one congruent with T which you labeled U to the right. 
Now you may be wondering that you don't know exactly how to position line T in your drawing. That is ok. Your not ready to put it in yet. 
On D you are going to construct a triangle with the left side of the trapezoid L used for the left side of the triangle and the right side of the trapezoid R used as the right side of the triangle. This is a problem of constructing a triangle with 3 known sides. 
Now that the triangle is drawn,  from the top of your triangle to the right endpoint  of line B (which is the same thing as the far right of line segment U) draw a line. This line is going to be a diagonal X. The idea is that you will be making a parallelogram and having diagonal X in place will simplify that. 
Now on diagonal X you draw a triangle given 3 known sides. Diagonal X will be the baseline. The triangle will be above and to the right of the diagonal. The two lateral sides of your triangle will be a side equal to T and a side equal to R. Start your line equivalent to line T from the upper left of diagonal X. Start your line equivalent to line R to the bottom right of diagonal X. 
Notice that the 2 triangles on each side of diagonal X are congruent because the 3 sides are congruent. Label congruent corresponding  angles. Notice which lines are implied to be parallel because of these angles. 
Notice which lines in your construction are equal. 
You should have your trapezoid as desired given 4 lines. 
If drawn correctly you should have the trapezoid you desire. 
Note there was nothing special about starting from the side that we did and drawing the triangle on the left. You could have just as easily worked the problem marking off a length equal to the upper baseline to the left on the bottom baseline and with drawing the first triangle to the right. 
Notice that we essentially cut the horizontal bit out of the trapezoid and pushed the left ramp going up to meet the right ramp going down. This created a triangle. The horizontal length of the upper base subtracted away from the lower base gave the base length of the triangle. Drawing triangles with 3 sides given....puts us back on familiar territory. We created a triangle then shoved the two triangle pieces back apart and added that horizontal part back into the middle. We used a parallelogram to make sure the lengths and orientations (angles) stayed correct. 
A: Translate the fourth side on the first. With this new center and radius the trapezoid third side draw again circle. (In figure circle with center G has radius BF)

