The ABC triangle is isosceles. AB = BC = 13, AC = 10. Find the distance from the vertex B to the point O of the intersection of the bisectors.
In the triangle ABC, the lateral sides AB + BC = 13 cm; base AC = 10 cm;
BН – bisector, height and median of triangle ABC;
BH² = AB² – AH²; AH = 1/2 AC = 1/2 * 10 = 5 (cm²); BH² = 13² – 5² = 169 – 25 = 144 (cm²);
BH = √144 = 12 (cm);
The median at the point of intersection is divided into segments in a ratio of 2: 1, counting from the top, that is:
RH: BH = 2: 1; ОВ = 2/3 ВН = 2/3 * 12 = 8 (cm);
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