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if $\theta$ is the angle, $b$ and $c$ are two lengths (36 and 59) and $a$ the length opposite $\theta$ (39) $$\cos(\theta)=\frac{b^2+c^2-a^2}{2bc}$$ $$\cos(\theta)=\frac{36^2+59^2-39^2}{2\times36\ti …

Consider the triangle CDA. We know the length AC as it is also a radius of the circle. Thus, we know the length of all three sides so we can find angle CAD with the cos rule. Let angle CAD = $\thet …

To illustrate @quasi's excellent answer: $A$ makes the lower cut, in red $B$ makes the blue bisector to the lower cut, crossing the midpoint, $D$ $A$ makes the upper, red cut $B$ makes the upper blu …

lazy hint: Move $\triangle CHD$ to the left of $\triangle ABG$. Move $\triangle BFG$ below $\triangle AGE$. Move $\triangle FHC$ below $\triangle EHD$. Count up your new rhombuses and decide what to …

Let $M$ be the midpoint of $\overline{AB}$. As $C,G$ and $A$ are collinear, $C,X$ and $M$ are collinear and $\overline{GX}$ is parallel to $\overline{AM}$, $\triangle CXG$ must be similar to $\triangl …

if a circle with centre $(x_s,y_s)$ and radius $r_s$ touches three circle with centres $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ and radii $r_1, r_2$ and $r_3$ respectively, then $(x_s-x_1)^2+(y_s-y_1)^2 …

But the sum of two sides of a triangle should not be longer than the third side, due to the "triangle inequality", which we ordinarily take as a given in geometry, or which we can prove algebraically ( …

$\sin(x)$, $\cos(x)$ and $\tan(x)$ are defined by the ratios of specific sides of right angled triangles. $$\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}},\quad\cos(A)=\frac{\text{adjacent}}{\te …

Your points lie on the midpoints of the sloped edges of the pyramid in a way that looks something like this. $\quad\quad\quad\quad\quad\quad\quad$$\tag{Face-on View}$ $\quad\quad\quad\quad\quad\quad …

Define the circle with centre $B$ as $c_B:\:(x-B_x)^2+(y-B_y)^2=BC^2$ and the circle with centre $D$ as $c_D:\:(x-D_x)^2+(y-D_y)^2=DC^2$. Find the intersection of $c_B$ and $c_D$ to get $C$ and the ot …

This is one reason why it is important for the axioms of Euclidean geometry to be well formulated. … Hilbert's, Tarski's, etc. formulations of geometry rectify some of the missing pieces that Euclidean geometry was missing; namely the uncertainty of whether the parallel postulate could be proven from …

We construct $\overline{CE}$ such that $E$ lies on $\overline{AB}$ and $\overline{DE}\,||\,\overline{CB}$. We then construct a circle through $A$, $B$ and $C$. Circle $ABC$ has a centre $O$ at the int …

This is a justification for the use of the rectangle but it might not be quite as rigorous as you're hoping for. The arc length of the circle at a given radius, $r$, is double the arc length of $\sqr …

This is a good way to start any geometry problem. You're given a rectangle, so draw the rectangle and label an unknown side-length $x$. …

Elaborating on the answers by Automatically Generated and Claude Leibovici, despite quintics not being generally solvable in radicals, we can obtain a closed form solution by permitting special functi …