AC' should be CA'

*Suppose that an angle, its bisector, and one side of this angle in one triangle are respectively congruent to an angle, its bisector, and one side of this angle in another triangle. Prove that such triangles are congruent.*

Hello there, would anyone care enough to parse this problem for me? I can't understand what the hypotheses mean, probably because English is not my native language. Thank you!

Edit: I'm sorry for posting, I figured out what my problem was. In the Czech Republic we are taught the bisector of an angle as a line so it did not come to me naturally that it's meant as a segment in this problem (and indeed it's always regarded as a segment in triangles).

Exercise 75: *Prove that a triangle that has two congruent angles is isoceles.*

$\S 40(3)$ - SSS-test proof establishes that in congruent triangles congruent angles are opposed to congruent sides. The given triangle is self-congruent, $\S 3$. Let the congruent angles be $\angle{1}$ and $\angle{2}$. Using the sides opposite congruent angles are congruent fact, the side opposite $\angle{1}$ is congruent to the side opposite $\angle{2}$. Therefore this triangle has 2 congruent sides and is isoceles, $\S 33$.

Exercise 76: *In a given triangle, an altitude is a bisector. Prove that the triangle is isoceles.*

The bisector has divided the angle at the vertex into halves and by being an altitude makes a right angle with the base, $\S 34$. The given triangle has been subdivided into two triangles that each have a common side (the altitude), with respectively congruent angles at both ends (half the vertex and a right angle). $\S 40(2)$ - ASA-test states that these two sub triangles are congruent. It follows that the sides opposite their right angles are congruent. Therefore the given triangle has 2 congruent sides and is isoceles, $\S 33$.

Exercise 77: *In a given triangle, an altitude is a median. Prove that the triangle is isoceles.*

$\S 40(1)$ SAS-test. The altitude has subdivided the given triangle into 2 sub-triangles. These sub-triangles share a common side (the altitude), which has a respectively congruent angle at the end (right; $\S 34$ altitude), and in both sub-triangles the other side of said angle is a mutually congruent side ($\frac{1}{2}$ the base; median $\S 34$). These sub-triangles are congruent and their hypotenuses are congruent. Therefore the given triangle has 2 congruent sides and is isoceles. $\S 33$.

Exercise 78: *On each side of an equilateral triangle $ABC$, congruent segments $AB'$, $BC'$, and $AC'$ are marked, and the points $A'$, $B'$, and $C'$ are connected by lines. Prove that the triangle $A'B'C'$ is also equilateral.*

**AC' is wrong for the marked congruent segment, it should be $CA'$ per Sumizdat 18 Jan 2009, 11:33**

The construction of $\triangle A'B'C'$ creates 3 other triangles surrounding it on the interior of $\triangle ABC$, $\triangle AB'A'$, $\triangle A'C'C$, and $\triangle B'BC'$. These 3 other triangles have a mutually congruent side (one of the marked congruent segments), another mutually congruent side (one of the complement segments to the marked congruent segments; $\overline{B'B} = \overline{C'C} = \overline{A'A}$), and these sides enclose a mutually congruent angle (#70 - Equilateral triangle $\triangle ABC$ is equiangular; $\angle{A} = \angle{B} = \angle{C}$). By $\S 40(1)$ SAS-test, these 3 other triangles are congruent and the remaining side in each that forms $\triangle A'B'C'$ are mutually congruent. Thus all sides of $\triangle A'B'C'$ are congruent and it is an equilateral triangle, $\S 33$.

Exercise 79: *Suppose that an angle, its bisector, and one side of this angle in on triangle are respectively congruent to an angle, its bisector, and one side of this angle in another triangle. Prove that such triangles are congruent.*

Let the first triangle be $\triangle ABC$, its mutually congruent angle be $\angle{C}$, the mutually congruent angle be the vertex of this triangle, and the point of intersection of the bisector of $\angle{C}$ and the base be point $X$.

Let the second triangle be $\triangle DEF$, its mutually congruent angle be $\angle{F}$, the mutually congruent angle be the vertex of this triangle, and the point of intersection of the bisector of $\angle{F}$ and the base be point $Z$.

The bisectors in both triangles create sub-triangles $\triangle ACX$ and $\triangle DFZ$ respectively. $\overline{CA} = \overline{FD}$, $\overline{CX} = \overline{FZ}$, and $\frac{1}{2}\angle{C} = \frac{1}{2}\angle{F}$. By $\S 40(1)$ SAS-test, these sub-triangles are congruent. It follows that the angles at the other end of the mutually congruent given sides are congruent. Therefore the given triangles have the respectively congruent given angle ($\angle{C} = \angle{F}$), the given side of this angle ($\overline{CA} = \overline{FD}$), and the angles at the other end of this given side ($\angle{A} = \angle{D}$). By $\S 40(2)$ ASA-test, the given triangles are congruent.

Exercise 80: *Prove that if two sides and the median drawn to the first of them in one triangle are respectively congruent to two sides and the median drawn to the first of them in another triangle, then such triangles are congruent.*

In each of the given triangles the respectively congruent two sides and median form sub-triangles that are congruent by $\S 40(3)$ SSS-test: half of base (median $\S 34$), median itself, remaining congruent side. From this it is known that the angles between the two respectively congruent sides of each of the given triangles are mutually congruent. By $\S 40(1)$ SAS-test, the given triangles are congruent.

Exercise 81: *Give an example of two non-congruent triangles such that two sides and one angle of one triangle are respectively congruent to two sides and one angle of the other triangle.*

The two congruent sides in each triangle must be deliberately selected to prevent the triangles from being congruent.

1. For the triangles to not be congruent, the mutually congruent sides must not both be sides of the mutually congruent angle, $\S 40(1)$ SAS-test. It follows that one of the two respectiely congruent sides in each triangle must be the side opposite the mutually congruent angle, and the two respectively congruent sides in each triangle are the sides of one of the other angles. E.g. $\angle{A}$ and $\angle{D}$ here:2. The sides in each triangle that belong to the respectively congruent angles cannot be respectively congruent. Let the two triangles be $\triangle ABC$ and $\triangle DEF$, and $\angle{C} = \angle{F}$. Suppose that a side of the mutually congruent angle is mutually congruent, $\overline{CA} = \overline{FD}$. This implies that the sides opposite these angles are also congruent, $\overline{AB} = \overline{DE}$. Juxtapose $\triangle ABC$ and $\triangle DEF$ such that point $C$ is taken to point $F$, $\overline{CB}$ is directed along $\overline{FE}$, $\angle{C}$ is adjacent to $\angle{F}$, and $\angle{A}$ and $\angle{D}$ are on opposite sides of $\overline{CB}$. Join $\angle{A}$ and $\angle{D}$ with a straight line to form $\triangle ACD$ with congruent sides $\overline{CA} = \overline{CD}$, and congruent base angles $\angle{CAD} = \angle{CDA}$ per $\S 35(2)$.

Isoceles $\triangle ABD$ has also been created with congruent sides $\overline{BA} = \overline{BD}$, and congruent base angles $\angle{BAD} = \angle{BDA}$. $\angle{A}$ and $\angle{D}$ are both the sum of mutually congruent angles and are thus congruent.

$\triangle ABC$ and $\triangle DEF$ are congruent by the $\S 40(1)$ SAS-test.Two non-congruent triangles that meet the constraints:

Exercise 82: *On one side of an angle $A$, the segments $AB$ and $AC$ are marked, and on the other side the segments $AB' = AB$ and $AC' = AC$. Prove that the lines $BC'$ and $B'C$ meet on the bisector of the angle $A$.*

Let the intersection of $\overline{BC'}$ and $\overline{B'C}$ be point $D$. With the given information the triangles $\triangle CAB'$ and $\triangle C'AB$ are known. $\overline{AC} = \overline{AC'}$, $\angle{A} = \angle{A}$, and $\overline{AB} = \overline{AB'}$. By $\S 40(1)$ SAS-test, these triangles are congruent. It follows that $\angle{C} = \angle{C'}$, $\angle{ABC'} = \angle{AB'C}$, and the supplementary angles ($\S 22$) to the prior set $\angle{CBC'} = \angle{C'B'C}$.

The given information has also produced triangles $\triangle CBD$ and $\triangle C'B'D$. From the proceeding congruencies it is known that $\overline{BC} = \overline{B'C'}$, $\angle{CBD} = \angle{C'B'D}$, and $\angle{C} = \angle{C'}$. $\S 40(2)$ ASA-test, $\triangle CBD = \triangle C'B'D$ and $\overline{BD} = \overline{B'D}$.

Join $B$ and $B'$ with a straight line. $\triangle BAB'$ and $\triangle BDB'$ both have two congruent sides and are isoceles, $\S 33$. The bisector of $\angle{A}$ is also the altitude and median of $\triangle BAB'$ ($\S 35(1)$, $\S 34$). The bisector of $\triangle BDB'$ is also an altitude and median. Thus the bisectors of $\triangle BAB'$ and $\triangle BDB'$ share the midpoint of $\overline{BB'}$, make right angles to said segment from opposing sides, and create a straight angle ($\S 16$). Point $D$ is on the bisector of $\angle{A}$.