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Q: Can the angle bisector of a triangle also be the perpendicular bisector?

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True

iscoceles triangle! =)

No. The angle bisector is a line. Where the three lines meet is the median. In an equilateral triangle the INTERSECTION of the angle bisectors is the median.

the circumcenter, orthocenter, and centriod, when connected together i Euler's line. the angle bisector of the non base angle is the same thing.

a bisector splits an angle into to congruent angles(also called an angle bisector)

Related questions

any isosceles triangle

True

Every isosceles or equilateral triangle.

iscoceles triangle! =)

Yes - the altitude of an equilateral triangle is perpendicular to the side chosen as the base and bisects that side and the opposite angle. Also, the altitude of an isosceles triangle when measured from the third side (the side that is not equal to the other two sides) is a perpendicular bisector of the base and also bisects the opposite angle.

-- An isosceles triangle has two equal sides. -- An isosceles triangle has two equal angles. -- An isosceles triangle has two equal interior-angle bisectors. -- The bisector of the vertex angle of an isosceles triangle is also the perpendicular bisector of the triangle's base.

Yes, provided that the base is not one of the 2 equal sides. And it's also the perpendicular bisector of the base.

The three ANGLE bisectors of a triangle also bisect the sides, and intersect at a point INSIDE the triangle. The angle bisectors are not necessarily perpendicular to them. The perpendicular bisectors of the sides can intersect in a point either inside or outside the triangle, depending on the shape of the triangle.

No. The angle bisector is a line. Where the three lines meet is the median. In an equilateral triangle the INTERSECTION of the angle bisectors is the median.

An isosceles or an equilateral triangle perhaps?

Label the triangle ABC. Draw the bisector of angle A to meet BC at D. Then in triangles ABD and ACD, angle ABD = angle ACD (equiangular triangle) angle BAD = angle CAD (AD is angle bisector) so angle ADB = angle ACD (third angle of triangles). Also AD is common. So, by ASA, triangle ABD is congruent to triangle ACD and therefore AB = AC. By drawing the bisector of angle B, it can be shown that AB = BC. Therefore, AB = BC = AC ie the triangle is equilateral.

The Angle Bisector Theorem states that given triangle and angle bisector AD, where D is on side BC, then . Likewise, the converse is also true. Not sure if this is what you want?

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