Geometry problems- really need help solving?

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Problem 1 Use a compass and a ruler to draw the angle BAC = angle A. Take any segment MR as a unit On the sides of the angle we plot AK = 3 MR and AE = 4MP. We construct the triplet ACE similar to triplet ABC by the second similarity sign 2 I use a compass and a ruler to draw the medians of ACE and find the distance from the intersection of the medians O to the vertex E, thus I find the segment OE. Let the distance from the intersection point of the medians D of the triangle ABC to the vertex C be the segment DC, it is given in the problem statement 3 On the side we construct an arbitrary angle and on one side of it I measure OE, AC and AE one after another from the vertex, and on the other side, I also measure the given segment DC from the vertex. Connect the end of segment OE with the end of segment DS and draw straight lines through the ends of segments on the first side of the angle. Then on the second side of the angle I obtain two more segments CT and TF 4 I construct the given angle A and on its sides from the vertex we draw segments equal to CT and TF The ends of these segments join. Let BD=x, then AD=13 -x Then the height of DC is the average of these segments, i.e. DC²= x 13 -x or 36= 13x - x² or x² -13x +36=0 wherefore x= 9 or 4 Consider the variant x=4 variant x=9 is symmetrical to the considered one Then BD=4 and AD=13-4=9 Each cathetus is the average proportional to the hypotenuse and the adjoining segment of hypotenuse, therefore 1 AC²=13*9= 117 or AC=√117= 3√13 2 BC²=13*4= 52 or BC=√52= 2√13