# Prove that two triangles are congruent if a median and the angles between the median and the including Sides of the one are respectively congruent to the median and the angles between it and the including sides of the other.

Mbalisikerc 2022-07-26 Answered
Prove that two triangles are congruent if a median and the angles between the median and theincluding Sides of the one are respectively congruent to the median and the angles between it andthe including sides of the other.
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## Answers (1)

Bradley Sherman
Answered 2022-07-27 Author has 17 answers
A triangle having two sides of equal length is called an isosceles triangle. The two angles adjacent to the third side of an isosceles triangle are equal. If a triangle has two equal angles, it is an isosceles triangle, i.e. it has two equal sides too.
In an isosceles triangle, the median, bisectrix, and height of the vertex between the equal sides all coincide. This median/bisectrix/height divides the isosceles triangle into two congruent right triangles. If a median coincides with a height, or if a bisectrix coincides with a height, or if a median coincides with a bisectrix), then the triangle is isosceles - the adjacent sides are equal to each other..
If BD is the median and the height of the triangle ABC, then triangles ABD and CBD are equal because they are right triangles whose both legs are equal, respectively. Therefore, AB = BC. If BD is a bisectrix and is also a height, then triangles ABD and CBD are equal because they are right triangles with one common leg and equal acute angles. If BD is a median and a bisectrix of triangle ABC, then triangles ABD and CBD are equal because all three angles of ABD and CBD are equal, respectively, and AD = CD.
On to the problem, in order to compare congruency between to different triangles we need to remember that two triangles are congruent (equal) if they have identical size and shape so that they can be exactly superimposed. Two congruent triangles ABC and CDA with a common side and equal angles located at two different ends of the common side always form a parallelogram.
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