#m_a

[coldhe05]

[marioma-a]

حلويات العيد، كحك و غريبة.

عيد فطر مبارك✨🤩

Ghoraiba and Kahk, Eid sweets

Eid el-Fitr✨

Eid Mubarak 😊💜

Photographed by me✌🏻

_____________________________

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[spr1ng-b0y]

I don't have any concrete LGBTQ+ hcs about fighters, but the only thing I know for sure is that there's nothing straight about ARMS at all

The whole title is mega gay

YEAAA haha!! This is what I've been saying lol!! You can't give your main characters dyed hair and pronouns and expect fans to believe they're heterosexual and cis 😭😭

[squareallworthy]

It's Triangle Tuesday! A triangle, its centroid, a theorem, and two bonus triangles.

Greetings, seekers of triangle knowledge. Today I'm going to talk about the centroid.

Among triangle centers, the centroid is, at first glance, maybe a bit underwhelming. Take a triangle ABC. Let the midpoints of the sides be M_a, M_b, and M_c. Draw a line from each vertex to the opposite midpoint. The three lines cross at a point G. And that's the centroid.

Very simple. Simple almost to the point of banality. Don't get me wrong - the centroid is a great point. It's just not very flashy in its construction, and the fact that it exists doesn't seem at all deep. Later on, we can show how the centroid relates to other things, such as the symmedian point, the Euler line, and the nine-point circle. But for now, it's kind of hard to appreciate the centroid on it own. Let's do what we can do now, before getting into that other stuff, to see why the centroid is interesting.

It's not at all difficult to prove that the lines meet at one point, but the proofs I've looked at aren't all that enlightening. You draw a line and look at some parallel lines, and you say, "mmm, okay, that checks out, that's a proof all right," but it doesn't feel like you've learned much.

Even before proving that the lines are all coincident, it seems obvious that they should be. Each line cuts the triangle into two equal parts, and so when you draw two of them, the point where they cross is the middle of the triangle, measured two ways. So it just makes sense that the third line has to go through the same point, right?

Let's see if we can follow our intuition on this to get at something more fundamental. The lines that connect the vertices to the midpoints of the side are called medians. In general, lines like this that extend from a vertex to the opposite side of the triangle are called cevians (pronounced ˈtʃeviən, CHAY-vee-un). Are the medians really all that special among other cevians? What if instead of marking the midpoint of each side, dividing it into two segments in a 1:1 ratio, we mark a point that divides the side in a 1:3 ratio, going counterclockwise around the triangle? If we then draw the cevians, what do they look like?

Not surprisingly, the they aren't coincident. They outline a triangle (in red) that looks like the original triangle, but smaller and tilted counterclockwise. If we divided in a 1:3 ratio going clockwise, we would get a clockwise-twisted triangle. And if we move the marked points back and forth, sure enough, only when the points are halfway along the sides do the three cevians cross at one point.

And if we take the crossing point of two cevians, and then draw the third line through that point? Where does that hit the opposite side? Here I'll go back and draw that with a dashed line on the 1:3 ratio counterclockwise drawing:

It lands pretty close to the vertex. Somehow that point has to balance the 1:3 ratio we used to measure the other points, but it's not clear exactly how.

So how are coincident points connected to the place where we cut the sides? Let's get a bit more formal. Let's have a triangle ABC, and a point P that is not located on any of the sides of the triangle. We'll draw cevians from the vertices through P, and they will cross the sides of the triangle at D, E, and F.

So we would like to know where F, for instance, cuts line AB, and how it relates to the other points. Or, to put it another way, if we knew the ratio of the length of AF to FB, and the same for BD to DC, could we say what the ratio of CE to EA is? Conversely, if we have a triangle with three cevians cutting the sides at D, E, and F, and we knew all the ratios AF/FB, BD/DC, and CE/EA, could we say whether the cevians are coincident or not?

It's easy to get lost here with statements made up of long strings of segments. If you are like me, your eyes start to glaze over when you see that and you don't learn anything. So let's lay out a plan for figuring this out. We are looking for some information about ratios of line segments. To do that, it would be helpful to have some similar triangles, because similar triangles have all the same angles, and differ only in size and orientation. So if you know something about the ratio of two sides of a triangle, you know the same thing about the corresponding sides of a similar triangle.

And a good way to get some similar triangles is to arrange to have them meet vertex-to-vertex between parallel lines, like this:

With this arrangement and a little bit of Euclid (which I won't get into here), we can show that the pairs of angles marked with the same letters are equal. The two triangles with angles α, β, and γ are therefore similar, and we can say that the corresponding sides are in the same proportions -- that is, if we compare the red dashed segment to the blue dashed segment, it is the same ratio as the red solid segment to the blue solid segment. The four colored segments together make a Z-shaped figure, and it is this kind of arrangement of segments that we want to consider as we figure out what's going on with our cevians.

So with that in mind, let's go back to triangle ABC and add a line through A that is parallel to side BC. The new line meets two of our cevians at G and H.

And that creates sets of similar triangles:

AHF is similar to BCF (in red, below)

AEG is similar to BCE (green)

AGP is similar to BDP (blue)

AHP is similar to CDP (yellow)

From that, we can get these relationships:

AF/FB = AH/BC (from the red triangles)

CE/EA = BC/AG (green)

AG/BD = AP/DP (blue)

AH/DC = AP/DP (yellow)

We are interested in the ratios that the sides of ABC are divided into, that is, AF/FB and so on. We have two of them above, which I have bolded. We still need to get BD/DC and then shuffle things around to get all of those into one equation.

The two equations above from the blue and yellow triangles have the same right hand side, so we can say

AG/BD = AH/DC

and by reshuffling,

BD/DC = AG/AH.

There's the ratio for the third side. Now let's multiply that together with the equations we got from the red and green triangles:

AF/FB * CE/EA * BD/DC = AH/BC * BC/AG * AG/AH.

Everything on the right cancels out, so if we reorder the things on the left side to be alphabetically nice, we have

AF/FB * BD/DC * CE/EA = 1

which is the first half of what we wanted to know. If three cevians pass through the same point, then they cut the sides into ratios that multiply to 1. What a nice simple relationship to remember!

What about the converse? Can we prove that if three cevians cut the sides in ratios that multiply to 1, they all pass through the same point?

Let's suppose that we have divided the three sides of ABC with points C, D, and E in a way that AF/FB * CE/EA * BD/DC = 1. Then let's draw the cevians BE and CF and say that their crossing-point is called P.

Now, if we draw a cevian from A through P, does it land at D, as our equation says it should? Well, it has to land somewhere, and we'll call that point D'.

Can we prove that D' = D?

We were given that

AF/FB * BD/DC * CE/EA = 1.

And since we just drew three coincident cevians, we can conclude from what we just proved that

AF/FB * BD'/D'C * CE/EA = 1.

Combining those, we do a little algebraic manipulation:

BD/DC = BD'/D'C

BD/DC + 1 = BD'/D'C + 1

BD/DC + DC/DC = BD'/D'C + D'C/D'C

(BD+DC) / DC = (BD'+D'C) / D'C

But BD+DC is the whole side BC of the triangle, and the same with BD'+D'C. So

BC / DC = BC / D'C

and therefore D = D'. And that's our theorem:

In a triangle ABC, lines connecting the vertices to points on the opposite sides D, E, and F are concurrent if and only if AF/FB * BD/DC * CE/EA = 1

This is called Ceva's theorem, after the Italian mathematician Giovanni Ceva (ˈtʃeva, CHAY-va), who proved it in 1678. But as usual in mathematics the theorem is not named for the original discoverer. That was Abu Amir Yusuf ibn Ahmad ibn Hud, who proved it in the 11th century. (I got the above proof from this website, though I have cut some corners by ignoring signed distances and neglecting the case with an obtuse triangle.)

Given Ceva's theorem, we can trivially prove that the medians of a triangle are concurrent. The midpoints divide each side in the ratio 1/1, so

1/1 * 1/1 * 1/1 = 1

proves the existence of the centroid. There are much more direct ways to prove this, of course, but Ceva's theorem will come up over and over again in the study of triangles so it's worthwhile to get it down now.

Now since we've proved that the medians are coincident, here are some of the properties of the medians and centroid.

The medians divide the triangle into six triangles of equal area.

Each median divides the triangle into two triangles of equal area, because they have equal bases and heights. Therefore

T+U+V = X+Y+Z

but also

Z+T+U = V+X+Y

and subtracting the second from first gives

V-Z = Z-V

V = Z

and similarly for every pair of opposite triangles. But also, X and V have equal bases and heights, so X = V, similarly T = U, Y = Z. Putting that all together shows that all six areas are equal.

The centroid G divides the triangle ABC into three triangles ABG, BCG, and ACG of equal area.

This follows immediately from the previous result, and we can immediately follow that with

The centroid lies on each median at 2/3 of the distance from a vertex to a midpoint.

ABG has the same base as ABC, and 1/3 the area, so it must have 1/3 the height. Those two very straightforward facts give us this less obvious property:

A point X divides a triangle ABC into equal triangles ABX, BCX, and ACX only if X is the centroid.

For a triangle ABX to have 1/3 the area of ABC, it must have a height 1/3 that of ABC, so X must lie on the dotted line in the picture, which runs through G. Analogous lines for triangles BCX and ACX must also concur at G.

And, of course, the most famous properties of the centroid, and the reason G is traditionally used to name it, are that the centroid is the center of gravity of the vertices, and also of the area of the triangle. (If you are coming from a physics background, you might object that the centroid of an object is by definition its center of gravity, or center of mass -- that's simply what "centroid" means. To avoid any confusion, I am using a gemometric definition of centroid as the place where the medians cross, and proving that this point is also the center of gravity.)

The centroid is the center of gravity of the vertices.

If A and B represent equal masses M, they can be replaced by a single mass of 2M located at Z. The center of gravity of C and Z, then, is a point 1/3 of the way from Z to C, which is G.

The centroid is the center of gravity of the area of the triangle.

Every line parallel to AB can be replaced with a mass Z_1, Z_2, ... at the midpoint of the line. Each of these masses will be located on the median, and so the center of gravity of all these points must also lie on the median. By a similar argument, the center of mass must also line on the other medians, so it must lie at their crossing point. (Yes I admit this is rather handwavy, but a better proof requires integration and I am doing elementary geometry here, not calculus. I have an idea for a proof using purely geometrical methods but I haven't finished it.)

To be complete, I will note that the centroid is not the center of gravity of the perimeter of the triangle. That's a different triangle center, which we will perhaps consider on a different day.

Finally let me introduce two triangles related to the centroid and the medians. The medial triangle, shown in red, is the triangle formed by the midpoints of the sides. The antimedial triangle, shown in blue, formed by lines passing through the vertices A, B, and C and parallel to the opposite side, which meet at A', B', and C'. The anti- part comes from the fact that ABC is the medial triangle of A'B'C.

Both of these new triangles are similar to the reference triangle ABC and can be formed by reflecting ABC through a point and scaling up or down by a factor of 2. That point of reflection is, of course, the centroid.

If you found this interesting, please try drawing some of this stuff for yourself! You can use a compass and straightedge, or software such as Geogebra, which I used to make all my drawings. You can try it on the web here or download apps to run on your own computer here.

An index of all posts in this series is available here.

[matcha-aroma]

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[inmergersen]

What are the Benefits of Spin-offs? Explaining the Differences from Subsidiary Companies in Detail

Regardless of the company's size, spin-off are carried out to strengthen specialization and enable rapid business expansion.

In this article, we will outline the overview of spin-offs, organize the advantages and disadvantages, and provide detailed explanations of points to consider when conducting spin-offs.

See more at inmergers.com/en https://inmergers.com/en/spin-off

#Inmergers #mmatch #m_a #MergersandAcquisitions #BenefistofSpin_offs #Spin_offs

[hanimazenabukhait]

Israel vs H!@ a. m_a -s

[yinnebomacompany]

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[yinnebomaanimationstudios]

Chaotic Comics Animated cartoon series voice actor Lee Braithwaite voice actors test footage check it out right here on the official chaotic comics animated cartoon series YouTube channel: https://m.youtube.com/watch?v=xfPZuS2-m_A Voice Actor Lee Braithwaite as Character Butler Ben for Chaotic Comics Series Voice Actors Test Footage. Aeather Inc. Company and Heptagon Arts Studios LLC. Company Presents: Chaotic Comics Animated Cartoon Series is an animated musical comedy fantasy series about a group of friends that all passionate and love music and forms a band and get's discovered and goes on a world tour around the world and then space in the first ever intergalactic space grand universe multiverse tour playing music and getting into all kinds of adventures. Created By Creator/Animator/Cartoonist/Filmmaker Yinneboma Peterson Produced By Aeather Inc. and Heptagon Arts Studios LLC. #art #arts #artist #artwork #studio #studios #create #creative #creativity #creation #creations #creating #creator #creators #animation #animations #cartoons #cartoon #comic #comics #toon #toons #cartoonist #animator #filmmaking #filmmaker #film #films #company #business https://www.instagram.com/p/CqObd5xvLPx/?igshid=NGJjMDIxMWI=

[delraycomputers]

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[ninahanen]

ليكن رحيلك  مسبق الدفع  دفعة وجع واحدة لا أريد أن أموت بالتقسيط  المؤلم  جرعة من البعد القاتل تكفي .. فارحل بكل ما أوتيت من البعد

[coldhe05]

[marioma-a]

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[spr1ng-b0y]

Can you please do Genderfluid and Gay Kid Cobra please?

Sure thing!

Feel free to use with credit

[morrisalbert]

Cada momento de vida é um momento de gratidão. ✨🙏🏻✨ . . . #washington_dc #usa #trip #tbf #morris_albert #m_a #morrinhobr #martinlutherkingjr #abrahamlincoln #MonumentodeWashington (em Washington DC, USA) https://www.instagram.com/p/B7-En6MlAr8/?igshid=tztbe4v5ucjd

[matcha-aroma]

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