Problems

In the triangle \(ABC\) the median \(BD\) coincides with the height. Prove that \(AB=BC\).

In the triangle \(ABC\) with \(BC=12\), the median \(AE\) is perpendicular to the bisector \(BD\). Find the length of \(AB\).

On the sides of the equilateral triangle \(ABC\) three points \(D,E,F\) are chosen in such a way that the following ratios of lengths hold: \[AD:DC = CF:FB = BE:AE\] Prove that the triangle \(DEF\) is also equilateral.

Two circles with centres \(A\) and \(C\) intersect at the points \(B\) and \(D\). Prove that the segment \(AC\) is perpendicular to \(BD\). Moreover, prove that the segment \(AC\) divides \(BD\) in half.

For two congruent triangles Prove that their corresponding heights are equal.

The sides \(AB\) and \(CD\) of the quadrilateral \(ABCD\) are equal, the points \(E\) and \(F\) are the midpoints of \(AB\) and \(CD\) correspondingly. Prove that the perpendicular bisectors of the segments \(BC\), \(AD\), and \(EF\) intersect at one point.

In the triangle \(ABC\) the heights \(AD\) and \(CE\) intersect at the point \(F\). It is known that \(CF=AF\). Prove that the triangle \(ABC\) is isosceles.

In the triangle \(ABC\) the angle \(\angle ABC = 120^{\circ}\). The segments \(AF,\, BE\), and \(CD\) are the bisectors of the corresponding angles of the triangle \(ABC\). Prove that the angle \(\angle DEF = 90^{\circ}\).

In the triangle \(ABC\) the lines \(AE\) and \(CD\) are the bisectors of the angles \(\angle BAC\) and \(\angle BCA\), intersecting at the point \(I\). In the triangle \(BDE\) the lines \(DG\) and \(EF\) are the bisectors of the angles \(\angle BDE\) and \(\angle BED\), intersecting at the point \(H\). Prove that the points \(B,\,H,\, I\) are situated on one straight line.

In the triangle \(ABC\) the points \(D,E,F\) are chosen on the sides \(AB, BC, AC\) in such a way that \(\angle ADF = \angle BDE\), \(\angle AFD = \angle CFE\), \(\angle CEF = \angle BED\). Prove that the segments \(AE, BF, CD\) are the heights of the triangle \(ABC\).