Problems

Age
Difficulty
Found: 1974

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.
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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.
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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.
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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.
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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 ABC=120. The segments AF,BE, and CD are the bisectors of the corresponding angles of the triangle ABC. Prove that the angle DEF=90.
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In the triangle ABC the lines AE and CD are the bisectors of the angles BAC and BCA, intersecting at the point I. In the triangle BDE the lines DG and EF are the bisectors of the angles BDE and BED, intersecting at the point H. Prove that the points B,H,I are situated on one straight line.
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In the triangle ABC the points D,E,F are chosen on the sides AB,BC,AC in such a way that ADF=BDE, AFD=CFE, CEF=BED. Prove that the segments AE,BF,CD are the heights of the triangle ABC.
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