Is it possible to construct an isosceles triangle by using a ruler and without using a pair of compasses? It is well known that on Euclidean plane one can construct an isosceles triangle on given straight line by using a ruler and a pair of compasses.

Also it is possible to construct straight line containing given point and parallel to given straight line by using only a ruler, with the condition that we can measure out a segment equal to any given segment.

But in this conditions there are difficulties with constructing perpendicular for given straight line. Ofcourse it is equivalent to constructing of an isosceles triangle on given straight line by using only a ruler, with the condition that we can measure out a segment equal to any given segment.

I haven't yet able to do this, also I haven't able to prove that it is impossible. Do anybody know something about this topic?

Thanks in advance.
Yes, this is possible.

It is possible to construct a pair of perpendicular lines somewhere on plane. To do this, make a rhombus. (On two lines that meet at a point A, construct segments of the same length starting from A. Then constructing parallels yields a parallelogram with two adjacent sides of equal length. This is a rhombus.) The diagonals of the rhombus are perpendicular.

Think of the two perpendicular lines as the x-axis and the y-axis, so we may draw horizontal and vertical lines through any point. This allows us to assume that the origin $(0,0)$ is on the given line. Choose some other point $(x,y)$ on the line. The horizontal and the vertical lines through $(x,y)$ meet the two axes at $(x,0)$ and $(0,y)$. So we may construct the points $(y,0)$ and $(0,−x)$. The vertical line through $(y,0)$ meets the horizontal line through $(0,−x)$ at the point $(y,−x)$. The line joining $(0,0)$ and $(y,−x)$ is perpendicular for the line line joining $(0,0)$ and $(x, y)$.
Wait, @EvgenyKuznetsovs, why are you answering your own questions? You literally answered one day after you asked. Its not like you put any real research into this if you could find an answer so soon after asking - or perhaps you already knew it when you asked?  And youre also going to award best answer to yourself, on top of it?

If you have the ability to measure length then simply extend two congruent line segments from any given point you wish, with any angle between them you wish, then connect their other endpoints.  Voila, you have an isosceles triangle. Being able to measure length is immensely powerful.

This business about parallel and perpendicular lines and rhombi is a bit unnecessary and excessive.  Don't you think? You only asked if it was possible, not how to construct a particular one in a particular way. You put no other conditions or stipulations on this, so there you go.

By "ruler" did you mean straightedge? I ask because people frequently abuse language and speak inexactly because they are ignorant of the particular meanings.

If you mean straightedge then 'no', you cannot construct an isosceles triangle. Insofar as I am aware. Not without a compass anyway, or having at least one circle already existing in the plane, or a way of measuring out length. Im no expert, per se, but Ive been studying both restricted and liberated construction methods to Euclidean geometry for a number of years now.

The answer you provided yourself in your response to yourself relies on the ability to measure lengths, which is not possible with a straightedge. But as I pointed out, there is a more straightforward way of doing it than the construction you gave. Youre not doing a Euclidean construction; youre doing, for lack of a better term, a "Cartesian construction", since you've embedded length measure into the plane by way of a ruler.

If by "ruler" you meant an actual demarcated length-measuring device, or a straightedge you're allowed to mark, then 'yes', fairly trivially. As I have outlined above, in my second paragraph.  Your method works too, but is hyper-convoluted.  I say that because I presume there is no other context you left out, and all you wanted was some arbitrary isosceles triangle in the plane.

As for constructing parallel and perpendicular lines, your method is good, although unnecessary for the intended goal. In fact it would have taken me some time to realize how to do the perpendicular.

An alternative construction for the parallel can be found on the wikipedia article for the Poncelet-Steiner theorem (linked above). Therein is an animated GIF image depicting the construction step-by-step. Just ignore the circle; what you need are the two points on the line (the line from which youre making a parallel) that passes through the circle center. These three points can be found without a compass simply be making congruent end-to-end line segments with your length measuring device.