Elementary Geometry Word Problems Here!

I kept wondering of the best way to "see" why midpoint of $A=x_1,B=x_2$ is $(x_1+x_2)/2$.

What's the ratio between the segments $\frac{AF.BG}{FG}$ in the figure below?

"The sum of the squares of the diagonals is equal to the sum of the squares of the four sides of a parallelogram."
I find this property very useful while solving different problems on Quadrilaterals & Polygon,so I am very inquisitive about a intuitive proof of this property.

Show that the circumscribed circle passes through the middle of the segment determined by center of the incircle and the center of an excircle.

Find the area of a triangle with vertices $(0,1,1),(-<!--\; -\; -->1,-<!--\; -\; -->1,2),(2,3,1)$

Given a lattice $\mathrm{\Gamma <!--\; \Gamma \; -->}\subset <!--\; \subset \; -->\mathbb{C}$, a Theta function $\mathrm{\vartheta <!--\; \vartheta \; -->}:\mathbb{C}\to <!--\; \to \; -->\mathbb{C}$ is a holomorphic function with the following property:
$\mathrm{\vartheta <!--\; \vartheta \; -->}(z+\mathrm{\gamma <!--\; \gamma \; -->})=e^{2i\mathrm{\pi <!--\; \pi \; -->}{a}_{\mathrm{\gamma <!--\; \gamma \; -->}}z+b_{\mathrm{\gamma <!--\; \gamma \; -->}}}\mathrm{\vartheta <!--\; \vartheta \; -->}(z)$
for every $\mathrm{\gamma <!--\; \gamma \; -->}\in <!--\; \in \; -->\mathrm{\Gamma <!--\; \Gamma \; -->}$, and $a_{\mathrm{\gamma <!--\; \gamma \; -->}},b_{\mathrm{\gamma <!--\; \gamma \; -->}}\in <!--\; \in \; -->\mathbb{C}$.

Exercise: A Theta function never vanishes iff $\mathrm{\vartheta <!--\; \vartheta \; -->}(z)=e^{p(z)}$ with $p(z)$ a polynomial of degree at most 2.

Hint: The "only if" part is trivial. The hint is: show that $\log <!--\; \; -->(\mathrm{\vartheta <!--\; \vartheta \; -->}(z))=O(1+|z{|}^2)$. I tried to apply log on both sides, or derive one and two times, or everything I could have thought of. I don't get where the square comes from.

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?

Suppose I have 4 unit vectors in 3D and I know all the ${}^4{C}_2=6$ angles between them. These angles provide the complete description of this group of vectors. Now, I want to add anther unit vector to the mix. How many additional angles do I need to uniquely identify this new vector?

If K is the midpoint of AH, $P\in AB$, $Q\in AC$ and $K\in PQ$ such that $OK\perp <!--\; \perp \; -->PQ$ , then $OP=OQ$

A convex quadrilateral ABCD is inscribed and circumscribed. If the diagonals AC and BD are perpendicular, show that one of them divides the quadrilateral into two congruent right triangles.

Given: ABCD is a parallelogram,$\stackrel{¯<!--\; ¯\; -->}{AM},\stackrel{¯<!--\; ¯\; -->}{BN}$ angle bisectors,$DM=4\text{ft.}$, $MN=3\text{ft.}$Find: the perimeter of ABCD

Bertrand's postulate

claim :

for all $5<n\in <!--\; \in \; -->\mathbb{N}$ there is at least two prime numbers different in the opening interval $(n,2n)$.Use reinforcement of Bertrand's postulate in order to prove that :
for all $10<n\in <!--\; \in \; -->\mathbb{N}$ Exists that has at least two Prime factors in the factorization of $n!$ which appear with a power of 1.
Example : $n=11$ the primes 7,11 are such prime's meet the conditions.
But , for $n=10$ it dosen't meet the conditions because only 7 appears with power 1 in the factorization of 10!.

Hint : Consider two cases, $n$ even or $n$ odd

Attempt:
we need to use the claim in order to Implement the solution

Case (1): $n$ even
if $n$ is even we can rewrite $n=2t$
if we use the claim we can get $(2t,4t)$

Case (2): n odd
if $n$ is odd we can rewrite $n=2t+1$
if we use the claim we can get $(2t+1,2(2t+1))$

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.

Is there an elegant way to prove that the midpoint is $(\frac{x_1+y_1}{2},\frac{x_2+y_2}{2})$ using non-right angled similar triangles?

I came up with this question- how would you show 4 not equal to 6 (or m not equal to m+n ( n not 0)), using only Peano's Postulates?

I can see a number of things go wrong- for instance the Principle of Mathematical Induction seems to fail. Also possibly 0 seems to be in the image of the successor function in that case.

Let $A$,$B$ be two points on opposite sides of a line $l$. Then the line segment $AB$ intersects $l$.

My question is:
1. Using only the 4 postulates of Euclid, is there a way to make precise the meaning of ``opposite sides"?
2. Is the intersection guaranteed to exist using only the 4 postulates? If so, why is it true?

Let $I(f)={\int <!--\; \int \; -->}_a^{b}f(x)dx$ . The midpoint rule (open Newton-Cotes for $n=0$) is I0(f)=(b−a)f(a+b2)$I_0(f)=(b-<!--\; -\; -->a)f(\frac{a+b}{2})$
Show: For $f\in <!--\; \in \; -->C^1([a,b])$ holds $|I(f)-<!--\; -\; -->I_0(f)|\le <!--\; \le \; -->\frac{(b-<!--\; -\; -->a{)}^2}{4}\Vert <!--\; \Vert \; -->f^{\prime }{\Vert <!--\; \Vert \; -->}_{\mathrm{\infty <!--\; \infty \; -->}}$

How does $d_1$ equal $\frac{1}{2}\sqrt{{(x_2-<!--\; -\; -->x_1)}^2+{(y_2-<!--\; -\; -->y_1)}^2}$ ?

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.

Let K be a nonempty closed, convex subset of $R^d$ . Prove that if $x\notin <!--\; \notin \; -->K$ , then there exists a unique point $y\in <!--\; \in \; -->K$ that is closest to x.