Angle theorems Uncovered: Discover the Secrets with Plainmath's Expert Help

What's the value of the angle x in the figure below?
For reference: (Figure without scale ) $AK=8,JC=4,BE=10$ (Answer: ${53}^o)$)

$\mathrm{△<!--\; △\; -->}ABC$ has D a point on side BC. Points E and F are on sides AB and AC such that DE and DF are the angle bisectors of $\mathrm{\angle <!--\; \angle \; -->}ADB$ and $\mathrm{\angle <!--\; \angle \; -->}ADC$ respectively. Find the value of $\frac{AE}{EB}\ast <!--\; \ast \; -->\frac{BD}{DC}\ast <!--\; \ast \; -->\frac{CF}{FA}$

I am trying to prove the inscribed angle theorem using vectors. I fixed the dots $A=(\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->},\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})$, and $B=(\cos <!--\; \; -->\mathrm{\phi <!--\; \phi \; -->},\sin <!--\; \; -->\mathrm{\phi <!--\; \phi \; -->})$, and I took another point $C=(\cos <!--\; \; -->\mathrm{\psi <!--\; \psi \; -->},\sin <!--\; \; -->\mathrm{\psi <!--\; \psi \; -->})$ in the biggest arc AB.
My idea was to calculate ${\displaystyle \frac{⟨<!--\; ⟨\; -->A-<!--\; -\; -->C,B-<!--\; -\; -->C⟩<!--\; ⟩\; -->}{|<!--\; |\; -->A-<!--\; -\; -->B|<!--\; |\; -->|<!--\; |\; -->B-<!--\; -\; -->C|<!--\; |\; -->}}$, what according to my calculations is ${\displaystyle \frac{1+\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\mathrm{\phi <!--\; \phi \; -->})-<!--\; -\; -->\cos <!--\; \; -->(\mathrm{\psi <!--\; \psi \; -->}-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->})-<!--\; -\; -->\cos <!--\; \; -->(\mathrm{\phi <!--\; \phi \; -->}-<!--\; -\; -->\mathrm{\psi <!--\; \psi \; -->})}{2\sqrt{1+\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\mathrm{\psi <!--\; \psi \; -->})\cos <!--\; \; -->(\mathrm{\phi <!--\; \phi \; -->}-<!--\; -\; -->\mathrm{\psi <!--\; \psi \; -->})-<!--\; -\; -->\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\mathrm{\psi <!--\; \psi \; -->})-<!--\; -\; -->\cos <!--\; \; -->(\mathrm{\phi <!--\; \phi \; -->}-<!--\; -\; -->\mathrm{\psi <!--\; \psi \; -->})}}}.$.
My main difficulty here is the square root, which I can't get rid of. Does someone know how to proceed from here?Or maybe to solve the problem with vectors with a different approach?

O is intersection of diagonals of the square ABCD. If M and N are midpoints of OB and CD respectively ,then $\mathrm{\angle <!--\; \angle \; -->}ANM=?$

Hexagon ABCDEF has sides AB and DE of length 2, sides BC and EF of length 7, and sides CD and AF of length 11, and it is inscribed in a circle. Compute the diameter of the circle.
Why is $\stackrel{¯<!--\; ¯\; -->}{\mathit{A}\mathit{D}}$ the diameter of the circle?

Is there a formula for the solid angle at each vertex of tetrahedron?
A tetrahedron has four vertices as much as a triangle has three vertices. A tetrahedron therefore can have four solid angles as much as a triangle can have three angles.
Is there a general formula for the solid angle at each vertex of tetrahedron, if the length of each of the six edges are given? As much as one can use the law of cosines to determine the angle at each vertex of a triangle, as long as the lengths of each sides of triangle is given?

In $\mathrm{\Delta <!--\; \Delta \; -->}ABC,$, K and L are points on BC. AL is the bisector of $\mathrm{\angle <!--\; \angle \; -->}KAC$. $KL\times <!--\; \times \; -->BC=BK\times <!--\; \times \; -->CL$. Find $\mathrm{\angle <!--\; \angle \; -->}BAL$.

Find $\mathrm{\angle <!--\; \angle \; -->}ACB$ using the knowledge of cyclic quadrilaterals.

O is the center of the circle
Find $\mathrm{\angle <!--\; \angle \; -->}ACB$. It is given that $\mathrm{\angle <!--\; \angle \; -->}ABD={50}^{\circ <!--\; \circ \; -->}$ and $\mathrm{\angle <!--\; \angle \; -->}ACD={50}^{\circ <!--\; \circ \; -->}$ (inscribed angle theorem)
Can you help me here stating the reasons

Given a triangle $\mathrm{△<!--\; △\; -->}ABM$ such that $|AM|=|BM|$ and a point C such that the oriented angle $\mathrm{\angle <!--\; \angle \; -->}AMB$ has twice the size of $\mathrm{\angle <!--\; \angle \; -->}ACB$, show that $|CM|=|AM|$.
I am pretty sure that this must hold. Can somebody point me to an (elementary) proof?

Two circles intersect at C and D, and their common tangents intersect at T. CP and CQ are the tangents at C to the two circles; prove that CT bisects $\mathrm{\angle <!--\; \angle \; -->}PCQ$.

Geometry Proof: Convex Quadrilateral
A quadrilateral ABCD is formed from four distinct points (called the vertices), no three of which are collinear, and from the segments AB, CB, CD, and DA (called the sides), which have no intersections except at those endpoints labeled by the same letter. The notation for this quadrilateral is not unique- e.g., quadrilateral $ABCD=$ quadrilateral CBAD.
Two vertices that are endpoints of a side are called adjacent; otherwise the two vertices are called opposite. The remaining pair of segments AC and BD formed from the four points are called diagonals of the quadrilateral; they may or may not intersect at some fifth point. If X, Y, Z are the vertices of quadrilateral ABCD such that Y is adjacent to both X and Z, then angle XYZ is called an angle of the quadrilateral; if W is the fourth vertex, then angle XWZ and angle XYZ are called opposite angles.
The quadrilaterals of main interest are the convex ones. By definition, they are the quadrilaterals such that each pair of opposite sides, e.g., AB and CD, has the property that CD is contained in one of the half-planes bounded by the line through A and B, and AB is contained in one of the half-planes bounded by the line through C and D.
a) Using Pasch's theorem, prove that if one pair of opposite sides has this property, then so does the other pair of opposite sides.
b) Prove, using the crossbar theorem, that the following are equivalent:
1. The quadrilateral is convex.
2. Each vertex of the quadrilateral lies in the interior of the opposite angle.
3. The diagonals of the quadrilateral meet.

How is this angle measured in the triangle?
Given the image below:

Let the angle at vertex A be called $\mathrm{\alpha <!--\; \alpha \; -->}$. Polya says by looking at the diagram and the isoceles triangles $\mathrm{△<!--\; △\; -->}ACE$ and $\mathrm{△<!--\; △\; -->}ABD$ we can deduce that $\mathrm{\angle <!--\; \angle \; -->}DAE=\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{2}+90°$.
Where will I get $90°$ from?

In an isosceles triangle ABC with base AC, the exterior angle at vertex C is 123 degrees. Find the angle ABC. Answer in degrees

the perimeter of the rhombus is 40. and one of the angles is 30 degrees. find the area of ​​the rhombus

Is it given that two lines are perpendicular if a right angle is shown?
If I have a diagram, like the following:

Derive a relation between angles A,B and C.
Derive a relation between angles A,B and C (do not use other angles in the final relation):

Find the measure of the angle:
1. m<1 = ?
2. m<5 = ?
3. m <6 = ?

Proof using properties of an isosceles or right-angle triangle
Given a $\mathrm{△<!--\; △\; -->}ABC$ with sides $AB=BC$ and $\mathrm{\angle <!--\; \angle \; -->}B={100}^{\circ <!--\; \circ \; -->}$, prove that $a^3+b^3=3a^{2}b$ where $a=AB=BC$ and $b=AC$.

Let A and B be two different points in the plane. Find the Locus of all the points C so $\mathrm{\angle <!--\; \angle \; -->}ACB=\frac{2\mathrm{\pi <!--\; \pi \; -->}}{3}$.

The bisector of exterior angle of B intersects the bisector CF (F on AB) at Y. Determine the comparison of $YF:YC$ by considering a,b,c as the lengths of side of the triangle ABC.
What theorems should I use here? Or what theorems should I apply first?