Construct a triangle given an angle, the side opposite the angle and the median to the other side (researching the number of solutions is not required).

In a convex quadrilateral $ABCD$, $\angle ABC = 90^o$ , $\angle BAC = \angle CAD$, $AC = AD, DH$ is the alltitude of the triangle $ACD$. In what ratio does the line $BH$ divide the segment $CD$?

Inside the segment $AC$, an arbitrary point $B$ is selected and circles with diameters $AB$ and $BC$ are constructed. Points $M$ and $L$ are chosen on the circles (in one half-plane with respect to $AC$), respectively, so that $\angle MBA = \angle LBC$. Points $K$ and $F$ are marked, respectively, on rays $BM$ and $BL$ so that $BK = BC$ and $BF = AB$. Prove that points $M, K, F$ and $L$ lie on the same circle.

In triangle $ABC$, $M$ is the point of intersection of the medians, $O$ is the center of the inscribed circle, $A', B', C'$ are the touchpoints with the sides $BC, CA, AB$, respectively. Prove that if $CA'= AB$, then $OM$ and $AB$ are perpendicular. PS. There is a a typo

Given triangle $ABC$. Point $O_1$ is the center of the $BCDE$ rectangle, constructed so that the side $DE$ of the rectangle contains the vertex $A$ of the triangle. Points $O_2$ and $O_3$ are the centers of rectangles constructed in the same way on the sides $AC$ and $AB$, respectively. Prove that lines $AO_1, BO_2$ and $CO_3$ meet at one point.

A circle is located on the plane. What is the smallest number of lines you need to draw so that, symmetrically reflecting a given circle relative to these lines (in any order a finite number of times), it could cover any given point of the plane?