let ABC be a triangle with incenter I. Denote A1, the midpoint of BC and M’a the midpoint of arc BC containing vertex A. Prove

Triangle BIM is isosceles, so, MB = MI.

Given that I denote A1, which is the mid point of BC.

An isosceles triangle is a triangle that contains two equal sides. Sometimes it’s specified as having exactly two sides of equal length, and sometimes as having at least two sides of equal length.

The two equal sides are known as the legs and the third side is known as the base of the triangle.

The two angles opposite the legs are equal and are always acute, so the bracket of the triangle as acute, right, or blunt depends only on the angle between its two sides.

This is just angle chasing. Let A equal to angle BAC, B equal to angle CBA, C equal to angle ACB,and note that A, I, M are collinear (as L is on the angle bisector). We are going to show that MB = MI, the other cases being similar.

First, notice that

∠MBI = ∠MBC + ∠CBI = ∠MAC + ∠CBI = ∠IAC + ∠CBI = 12 A + 12 B.

However,

∠BIM = ∠BAI + ∠ABI =12 A + 12 B.

Hence, BIM is isosceles.

So, MB = MI.

Hence triangle BIM is isosceles, so, MB = MI.

Question is incomplete. The question should include:

An isosceles triangle is a triangle that contains two equal sides. Sometimes it’s specified as having exactly two sides of equal length, and sometimes as having at least two sides of equal length, the ultimate interpretation thus including the equilateral triangle as a special case.

Exemplifications of isosceles triangles include the isosceles right triangle, the golden triangle, and the faces of bipyramids and certain Catalan solids.

The fine study of isosceles triangles dates back to ancient Egyptian mathematics and sumptuous mathematics. Isosceles triangles have been used as decoration from indeed before times, and appear constantly in armature and design, for case in the pediments and gables of structures.

The two equal sides are known as the legs and the third side is known as the base of the triangle.

The other confines of the triangle, similar as its height, area, and border, can be calculated by simple formulas from the lengths of the legs and base. Every isosceles triangle has an axis of harmony along the vertical bisector of its base.

The two angles opposite the legs are equal and are always acute, so the bracket of the triangle as acute, right, or blunt depends only on the angle between its two sides.

This is just angle chasing. Let A equal to angle BAC, B equal to angle CBA, C equal to angle ACB,

and note that A, I, M are collinear (as L is on the angle bisector). We are going to show that MB = MI, the other cases being similar.

First, notice that

∠MBI = ∠MBC + ∠CBI = ∠MAC + ∠CBI = ∠IAC + ∠CBI = 12 A + 12 B.

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