The bisectors of triangle ABC meet at point O, with angle AOB = angle BOC = 110 degrees

The bisectors of triangle ABC meet at point O, with angle AOB = angle BOC = 110 degrees. a) prove that triangle ABC is isosceles and indicate its base. b) find the corners of the given triangle

1). Triangle BOС is equal to triangle AOB along the side according to the second sign of equality of triangles: <AOB = <BOC – by the condition of the problem;

<ABO = <ОВС – ОВ – bisector of angle В;

OV side – common;

Therefore AB = BC, that is, the triangle ABC is isosceles with the base AC.

2). <AOC = 360 ° – <AOB – <BOC;

<AOC = 360 ° – 110 ° – 110 ° = 140 °;

The AOC triangle is isosceles, since AO = OC from the equality of the AOB and BOС triangles;

Means <ОАС = <ОАС = 1/2 (180 ° – <AOC);

<ОАС = <ОАС = 1/2 (180 ° – 140 °) = 20 °;

<A = 2 * <ОАС, since AO is the bisector of angle A;

<A = 2 * 20 ° = 40 °;

<C = <A = 40 °;

<B = 180 ° – <A – <C;

<B = 180 ° – 2 * 40 ° = 100 °.