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.

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.

You can still ask an expert for help

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.

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.

asked 2022-06-08

Find the equation to the pair of angle bisectors of the pair of lines $(ax+by{)}^{2}=3(bx-ay{)}^{2}$.

Efforts:

$(ax+by{)}^{2}=3(bx-ay{)}^{2}$

After simplifying, I got:

${x}^{2}({a}^{2}-3{b}^{2})+8abxy+{y}^{2}({b}^{2}-3{a}^{2})=0$

Now, what should I do next?

Efforts:

$(ax+by{)}^{2}=3(bx-ay{)}^{2}$

After simplifying, I got:

${x}^{2}({a}^{2}-3{b}^{2})+8abxy+{y}^{2}({b}^{2}-3{a}^{2})=0$

Now, what should I do next?

asked 2022-06-04

I'm taking a course at teaching and we have some geometry questions. Among the questions there was one I couldn't solve.

I'm trying to prove that angle bisectors in a triangle intersect at a single point, but I need to show it analytically, i.e using line equations and distance between points. I'm familiar with proofs using geometry and vectors, but couldn't manage to show analytically. I can't use the formula of angle between lines or distance of point from a given line.

Things I've tried:

1. angle bisector theorem.

2. Writing equations of sides as $Ax+By+C=0$ and then creating the equation of angle bisector - quite messy and I believe there is a lot easier way.

Thanks!

I'm trying to prove that angle bisectors in a triangle intersect at a single point, but I need to show it analytically, i.e using line equations and distance between points. I'm familiar with proofs using geometry and vectors, but couldn't manage to show analytically. I can't use the formula of angle between lines or distance of point from a given line.

Things I've tried:

1. angle bisector theorem.

2. Writing equations of sides as $Ax+By+C=0$ and then creating the equation of angle bisector - quite messy and I believe there is a lot easier way.

Thanks!

asked 2022-06-24

I tried gooogling how to do it but there were none that could relate to the problem I have.

Find the angle bisectors of the lines:

$g:r=\left(\begin{array}{c}2\\ 5\\ -9\end{array}\right)+u\left(\begin{array}{c}8\\ 4\\ 1\end{array}\right)$

$f:r=\left(\begin{array}{c}2\\ 5\\ -9\end{array}\right)+v\left(\begin{array}{c}12\\ 4\\ 3\end{array}\right)$

Could someone give me an idea on how to start solving this question? I know that in order to determine the direction of the angle bisectors of the lines need to have the same length. So if

$u$ or $v$=0 then they would have the same length no? But idk how to go from there.

Find the angle bisectors of the lines:

$g:r=\left(\begin{array}{c}2\\ 5\\ -9\end{array}\right)+u\left(\begin{array}{c}8\\ 4\\ 1\end{array}\right)$

$f:r=\left(\begin{array}{c}2\\ 5\\ -9\end{array}\right)+v\left(\begin{array}{c}12\\ 4\\ 3\end{array}\right)$

Could someone give me an idea on how to start solving this question? I know that in order to determine the direction of the angle bisectors of the lines need to have the same length. So if

$u$ or $v$=0 then they would have the same length no? But idk how to go from there.

asked 2022-05-09

How does ${d}_{1}$ equal $\frac{1}{2}\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}$ ?

asked 2022-04-30

If a line L separates a parallelogram into two regions of equal areas, then L contains the point of intersection of the diagonals of the parallelogram.

The figure shows a line L horizontally through the sides of the parallelogram.

This creates two trapezoids and I can intuitively show that if the bases of the trapezoids are not congruent then the areas can not be equal.

I just can not currently see how to relate the intersection with the diagonals.

Any help will be appreciated.

The figure shows a line L horizontally through the sides of the parallelogram.

This creates two trapezoids and I can intuitively show that if the bases of the trapezoids are not congruent then the areas can not be equal.

I just can not currently see how to relate the intersection with the diagonals.

Any help will be appreciated.

asked 2022-06-13

In $\mathrm{\Delta}ABC$, $BE$ and $CF$ are the angular bisectors of $\mathrm{\angle}B$ and $\mathrm{\angle}C$ meeting at $I$. Prove that $AF/FI=AC/CI$

I have tried this question for hours but i am unable to hit the nut

I tried to using: (1)Similarity (2)Angle bisector theorem (3)Relation of lengths of angle bisector :- $B{E}^{2}+AE.EC=AB.BC$ $C{F}^{2}+AF.FB=AC.BC$

I tried the question with these many approaches but i unable to prove the above relation.please tell if any of these approach will help or please tell any other method to solve the question.

I have tried this question for hours but i am unable to hit the nut

I tried to using: (1)Similarity (2)Angle bisector theorem (3)Relation of lengths of angle bisector :- $B{E}^{2}+AE.EC=AB.BC$ $C{F}^{2}+AF.FB=AC.BC$

I tried the question with these many approaches but i unable to prove the above relation.please tell if any of these approach will help or please tell any other method to solve the question.

asked 2022-08-04

Choose 3 inequalities that form a system whose graph is the shaded region shown above.

$A.y\le -2\phantom{\rule{0ex}{0ex}}B.y\ge -2\phantom{\rule{0ex}{0ex}}C.7x-4y\ge -13\phantom{\rule{0ex}{0ex}}D.7x+2y\ge 17\phantom{\rule{0ex}{0ex}}E.7x-4y\le -13\phantom{\rule{0ex}{0ex}}F.7x+2y\le 17\phantom{\rule{0ex}{0ex}}G.x\ge -2\phantom{\rule{0ex}{0ex}}H.y\le 2$

$A.y\le -2\phantom{\rule{0ex}{0ex}}B.y\ge -2\phantom{\rule{0ex}{0ex}}C.7x-4y\ge -13\phantom{\rule{0ex}{0ex}}D.7x+2y\ge 17\phantom{\rule{0ex}{0ex}}E.7x-4y\le -13\phantom{\rule{0ex}{0ex}}F.7x+2y\le 17\phantom{\rule{0ex}{0ex}}G.x\ge -2\phantom{\rule{0ex}{0ex}}H.y\le 2$