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Prove that triangles $VAC$ and $VBD$ have equal areas and equal perimeters....

As commented by Blue, the claim is not true in general. In the following, I'm going to prove the following claim : Claim 1 : If $\angle AVC =\angle BVD\color{red}{\not=90^\circ}$, then triangles $VAC$ ...
mathlove's user avatar
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2 votes
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Angle between random vectors with uniformly distributed coordinates

For large $d$, this value will be close to the value you get if you substitute the expected value for each factor. The expected value of $X\cdot Y$ and of $\|X\|^2$ is in each case $n$ times the ...
joriki's user avatar
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1 vote
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Prove that the triangles $VAC, EAV$ are similar if and only if $\angle EVO=30°$.

Case 1: Well done; you could have just specified that $VC=VA(=2x\sqrt3)$ Case 2: Well done; you could have quoted the angle bissector theorem and perhaps write more classicaly $\frac{V\color{red}A}{V\...
Stéphane Jaouen's user avatar
1 vote
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Explanation to how the length between 2 centers is adjusted as angle changes

TL DR Yes, for any given angle you will need to adjust the x axys. Intuition Start with the two centers aligned on the x axys. If you move the center up the new distance between the two centers will ...
Marco's user avatar
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1 vote

Prove that triangles $VAC$ and $VBD$ have equal areas and equal perimeters....

Define $$a:=|AV| \quad b := |BV| \quad c := |CV| \quad d := |DV| \\[4pt] s := |AC|=|BD|$$ Note that the desired equal-perimeter property for $\triangle AVC$ and $\triangle BVD$ means $$a+c+s = b+d+s \...
Blue's user avatar
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1 vote

Prove that triangles $VAC$ and $VBD$ have equal areas and equal perimeters....

Let $A(0,0,0),$ $B(0,b,0),$ $C(a,b,0),$ $D(a,0,0)$ and $V(x,y,z)$. Then, the perimeter condition $VA+VD=VB+VC$ and the area condition $\Vert \vec{VA}\times\vec{VD}\Vert=\Vert\vec{VB}\times\vec{VC}\...
Bob Dobbs's user avatar
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1 vote
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Problem to prove a point lies on a circle.

Two versions for you: On the left, draw the circumcircle, start by forming the perpendicular bisector on $AC$, shown in gray. Then note that the angles marked by $1$ are the same since they subtend ...
RobinSparrow's user avatar
1 vote
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Same angles in complex plane

If you can show that $$ \arg\left(\frac{a-b}{c-b}\right) = \pm\arg\left(\frac{d-e}{f-e}\right) \tag1$$ then you have shown the angles are equal in magnitude. If you also care about the orientation of ...
David K's user avatar
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1 vote

How do we define exterior angle of a concave polygon whose interior is reflex?

Your $c^\circ$ is not an exterior angle of the concave quadrilateral (the kind of exterior angles that add to $360^\circ$). As illustrated on Wikipedia, the exterior angle at that reflex vertex is ...
peterwhy's user avatar
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