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Q: How do you write a two column proof that has triangle AbC is equiangular for the given and EF is parallel to AC and prove triangle EFB is equiangular?

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There are absolutely 0 parallel lines in any given triangle.

The length of a midsegment is half that of the parallel side of the triangle; assuming the midsegment is parallel to the [given] base, then its length is 27 ÷ 2 = 13.5 units.

A square would fit the given description

Given ABE, ADC, BD bisescts angle ABC and BD is parallel to EC prove: Triangle EBC is isoceles

Yes, a triangle can be constructed if all sides are given.

Related questions

There are absolutely 0 parallel lines in any given triangle.

Given a triangle and a line, if the line passes through two sides of the triangle parallel to the third side, then it cuts the sides proportionally.l

The length of a midsegment is half that of the parallel side of the triangle; assuming the midsegment is parallel to the [given] base, then its length is 27 ÷ 2 = 13.5 units.

A square would fit the given description

Given ABE, ADC, BD bisescts angle ABC and BD is parallel to EC prove: Triangle EBC is isoceles

The circumcircle of a triangle is the circle that passes through the three vertices. Its center is at the circumcenter, which is the point O, at which the perpendicular bisectors of the sides of the triangle are concurrent. Since our triangle ABC is an isosceles triangle, the perpendicular line to the base BC of the triangle passes through the vertex A, so that OA (the part of the bisector perpendicular line to BC) is a radius of the circle O. Since the tangent line at A is perpendicular to the radius OA, and the extension of OA is perpendicular to BC, then the given tangent line must be parallel to BC (because two or more lines are parallel if they are perpendicular to the same line).

The Playfair Axiom (or "Parallel Postulate")

A triangle can be constructed into any of the given formats.

Yes, a triangle can be constructed if all sides are given.

Euclid's parallel postulate.

Equilateral Triangle.

Equiangular Polygon...