Angle Bisectors Problems: Solutions & Examples

Let ABC be a triangle. Let B' and C' denote respectively the reflection of B and C in the internal angle bisector of $\mathrm{\angle <!--\; \angle \; -->}A$, Show that the triangle ABC and AB'C' have the same in centre?
What should be the approach in this question? I don't even know how to initiate.

Let $A_1,B_1,C_1$ be the tangency points of the intersection of the excircles of a triangle $ABC$ with the sides $BC,CA,AB,$ respectively. Prove that the circumcircles of $AB{B}_1$ and $AC{C}_1$ meet on a bisector of $\mathrm{\angle <!--\; \angle \; -->}BAC.$
What I thought: Circle $(AB{B}_1)$ has ${\mathrm{\omega <!--\; \omega \; -->}}_1=b(s-<!--\; -\; -->a)$ and circle $(AC{C}_1)$ has $v_2=c(s-<!--\; -\; -->a)$. Their radical axis is given by $-<!--\; -\; -->yc(s-<!--\; -\; -->a)+zb(s-<!--\; -\; -->a)$ which is equation of A angle bisector and we are done.

How to prove that angle bisector of right angle triangle ABC right angled at B is perpendicular bisector of third side AC

The exterior angle bisectors of $\mathrm{\angle <!--\; \angle \; -->}B$ and $\mathrm{\angle <!--\; \angle \; -->}C$ intersect on point $O$. $\mathrm{\angle <!--\; \angle \; -->}BOC=70°$.
Find $\mathrm{\angle <!--\; \angle \; -->}OAC$.

Let $I$ be the center of the inscribed sphere of a tetrahedron $ABCD$ and let $I_A$ be the length of the line passing through $I$ from the vertex $A$ to the opposite face. I am looking for a formula for $I_A^2$ which is analogous to the formula for the angle bisector in a triangle
$I_a^2=\frac{bc}{(b+c{)}^2}[(b+c{)}^2-<!--\; -\; -->a^2].$
I assume that if such a formula exist, it involves areas of the faces of the tetrahedron. Any comments are appreciated.

Given two different non-parallel vectors, $\stackrel{\to <!--\; \to \; -->}{a}$ and $\stackrel{\to <!--\; \to \; -->}{b}$, prove that $\stackrel{\to <!--\; \to \; -->}{a}$+$\stackrel{\to <!--\; \to \; -->}{b}$ and $\stackrel{\to <!--\; \to \; -->}{a}$−$\stackrel{\to <!--\; \to \; -->}{b}$ are the bisector vectors of the angle formed by $\stackrel{\to <!--\; \to \; -->}{a}$ and $\stackrel{\to <!--\; \to \; -->}{b}$.

This question is in my book of vector geometry, but it is kind of weird for me. I do not know if I am approaching it the right way.
I know that the formula for the angle bisector vector is given like this:
Let $u$ and $v$ be vectors of non-zero length. Let $\Vert <!--\; \Vert \; -->u\Vert <!--\; \Vert \; -->$ and $\Vert <!--\; \Vert \; -->v\Vert <!--\; \Vert \; -->$ be their respective lengths. Then $\Vert <!--\; \Vert \; -->u\Vert <!--\; \Vert \; -->v+\Vert <!--\; \Vert \; -->v\Vert <!--\; \Vert \; -->u$ is the angle bisector of $u$ and $v$. So the only way $\stackrel{\to <!--\; \to \; -->}{a}$+$\stackrel{\to <!--\; \to \; -->}{b}$ is the bisector vector is that both $\stackrel{\to <!--\; \to \; -->}{a}$ and $\stackrel{\to <!--\; \to \; -->}{b}$ have a length of one unit, but the question does not talk about this particular case, it seems phrased for all cases. Also I do not see how $\stackrel{\to <!--\; \to \; -->}{a}$ -$\stackrel{\to <!--\; \to \; -->}{b}$ could satisfy this formula.

In the triangle ABC, one has $\angle A={70}^o$.
The point $D$ is chosen on the segment $AC$ such that the angle bisector $AE$ intersects the segment $BD$ at the point $H$, with the following ratios
${\displaystyle \frac{AH}{HE}}={\displaystyle \frac{3}{1}},{\displaystyle \frac{BH}{HD}}={\displaystyle \frac{5}{3}}.$
Find the angle $\angle C$.

Triangle $ABC$; $AB=c,BC=a,AC=b$; angle bisector of angle ($c$,$a$) cuts $A$$C$ in point $E$
Why is the following true?
$|AE|=\frac{bc}{a+c}$
Where does that come from?

In each diagram, BD bisects

How do i prove that the three medians of a triangle are concurrent at a point called centroid?

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.

Internal angle bisector of $\mathrm{\angle <!--\; \angle \; -->}A$ of triangle $\mathrm{\Delta <!--\; \Delta \; -->}ABC$, meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at P and the side AB at Q. If a, b, c represent the sides of $\mathrm{\Delta <!--\; \Delta \; -->}ABC$ then.
(a) $AD=\frac{2bc}{b+c}\cos <!--\; \; -->\frac{A}{2}$
(b) $PQ=\frac{4bc}{b+c}\sin <!--\; \; -->\frac{A}{2}$
(c) The triangle $\mathrm{\Delta <!--\; \Delta \; -->}APQ$ is isosceles
(d) AP is HM of b and c
My approach is as follow

This is the rough image that I have drawn
$\frac{\cos <!--\; \; -->\frac{A}{2}}{AD}=\frac{\sin <!--\; \; -->{90}^o}{b_1}$
$\frac{\sin <!--\; \; -->X}{b_2}=\frac{\sin <!--\; \; -->C}{PD}$
$PD=AD\tan <!--\; \; -->\frac{A}{2}$
$\frac{\sin <!--\; \; -->X}{b_2}=\frac{\sin <!--\; \; -->C}{PD}\Rightarrow <!--\; \Rightarrow \; -->\frac{\sin <!--\; \; -->X}{b_2}=\frac{\sin <!--\; \; -->C}{AD\tan <!--\; \; -->\frac{A}{2}}$
The official answer is a,b,c and d.
I am not able to approach from here

Convert from radians to degrees:
$-<!--\; -\; -->\frac{3\mathrm{\pi <!--\; \pi \; -->}}{8}$

Convert from radians to degrees:
$\frac{5\mathrm{\pi <!--\; \pi \; -->}}{12}$

Convert from radians to degrees:
$\frac{4\mathrm{\pi <!--\; \pi \; -->}}{3}$

Convert from degrees to radians:
$-<!--\; -\; -->{315}^{\circ <!--\; \circ \; -->}$

Convert from degrees to radians:
${210}^{\circ <!--\; \circ \; -->}$

Convert from radians to degrees:
$-<!--\; -\; -->\frac{7\mathrm{\pi <!--\; \pi \; -->}}{2}$

Convert from degrees to radians:
$-<!--\; -\; -->{500}^{\circ <!--\; \circ \; -->}$

Convert from degrees to radians:
${300}^{\circ <!--\; \circ \; -->}$