# Tag Info

## Hot answers tagged angle

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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 $VAC$ ...
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Accepted

### 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 ...
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1 vote
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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\... • 1,971 1 vote Accepted ### 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 ... • 1,862 1 vote ### Prove that trianglesVAC$and$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 $$a+c+s = b+d+s \... • 75.3k 1 vote ### Prove that triangles VAC and VBD have equal areas and equal perimeters.... Let A(0,0,0), B(0,b,0), C(a,b,0), D(a,0,0) and V(x,y,z). Then, the perimeter condition VA+VD=VB+VC and the area condition \Vert \vec{VA}\times\vec{VD}\Vert=\Vert\vec{VB}\times\vec{VC}\... • 10.3k 1 vote Accepted ### Problem to prove a point lies on a circle. Two versions for you: On the left, draw the circumcircle, start by forming the perpendicular bisector on AC, shown in gray. Then note that the angles marked by 1 are the same since they subtend ... • 510 1 vote Accepted ### Same angles in complex plane If you can show that$$ \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 ... • 97.7k 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 ...
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