Angle Bisectors Problems: Solutions & Examples

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!

The angle bisectors $AL$ and $BQ$ in the right triangle $\mathrm{△<!--\; △\; -->}ABC$ with $\mathrm{\angle <!--\; \angle \; -->}ACB={90}^{\circ <!--\; \circ \; -->}$ and catheti 15 and 20 divide the median $CM$ into three segments. Find their lengths.

I am not sure how to approach the problem. The most "obvious" thing that we can do is to find the hypotenuse $c=\sqrt{a^2+b^2}=25$ of the triangle $\mathrm{△<!--\; △\; -->}ABC$. Therefore, $m_c={\displaystyle \frac{1}{2}}\cdot <!--\; \cdot \; -->25=12.5$. On my sketch $AC=15$and $BC=20$. What to do next?

In $\mathrm{△<!--\; △\; -->}ABC$, $BE$ and $CF$ are the angular bisectors of $\mathrm{\angle <!--\; \angle \; -->}B$ and $\mathrm{\angle <!--\; \angle \; -->}C$ meeting at $I$. Prove that $\frac{AF}{FI}=\frac{AC}{CI}$.
By the angle bisector theorem we have $\frac{AC}{AE}=\frac{CB}{BE}$ and $\frac{AB}{AF}=\frac{BC}{FC}$. How do I proceed after this? Hints would be appreciated.

I have the following problem: In a triangle ABC, the measure of the angle formed by the external angle bisectors of B and C is equal to twice the measure of the angle A. Find out the value of angle A.
Can anyone suggest how to approach this example?

The internal and external bisectors of angle $A$ of a triangle $ABC$ meet $BC$ (or the extension of $BC$) at $X$ and $X^{\prime }$ respectively, show that the circles $ABC$ and $AX{X}^{\prime }$ intersect orthogonally. I have shown that the centre of the circle $AX{X}^{\prime }$ lies on (line) $BC$, but cannot proceed further. Any advice would be appreciated.

Suppose you have a pair of lines passing through origin, $a{x}^2+2hxy+b{y}^2=0$, how would you find the equation of pair of angle bisectors for this pair of lines. I can do this for 2 separate lines, but I am not able to figure it out for a pair of lines. Can someone please help.

My son got this problem in geometry and was stumped. He asked me and I am stumped too. Here is the problem:

In triangle ABC, m∠ACB = 42°. The angle bisectors AD and BE intersect at point O so that AE + OE = AB. Find m∠A and m∠B.

The lines which bisect an angle and the adjacent angle made by producing one of its arms are called the internal and external bisectors of an angle.
What does "external angle bisector" mean? How an angle can be bisected externally?

We have a $\mathrm{△<!--\; △\; -->}ABC$ and a $\mathrm{△<!--\; △\; -->}{A}_1{B}_1{C}_1$. The segments $CL$ and $C_1{L}_1$ are angle bisectors. If $\mathrm{△<!--\; △\; -->}ALC\sim <!--\; \sim \; -->\mathrm{△<!--\; △\; -->}{A}_1{L}_1{C}_1$, I should show that $\mathrm{△<!--\; △\; -->}ABC\sim <!--\; \sim \; -->\mathrm{△<!--\; △\; -->}{A}_1{B}_1{C}_1$.

From the similarity, we have ${\displaystyle \frac{AL}{A_1{L}_1}}={\displaystyle \frac{CL}{C_1{L}_1}}={\displaystyle \frac{AC}{A_1{C}_1}}$. The only way I see from here is to show that $\mathrm{△<!--\; △\; -->}LBC\sim <!--\; \sim \; -->\mathrm{△<!--\; △\; -->}{L}_1{B}_1{C}_1$. Is this necessary for the solution?

How do I prove that a triangle with sides a, b, c, has an angle bisector (bisecting angle A) is of length:
$\frac{2\sqrt{bcs(s-<!--\; -\; -->a)}}{b+c}$
I have tried using the sine and cosine rule but have largely failed. A few times I have found a way but they are way too messy to work with.