How do you find the angle inside a kite?

Arts & Sciences Science & Nature
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Last thing I’ll post about this.

  1. Someone on reddit gave the following hints:

The solution is not particularly insightful, it mostly rests on some important results and ideas and you just have to join them together in a right way, so I’m not sure that a full solution will be that interesting or helpful

Ok, two hints may be really helpful: reflecting ABC across AB is nice and finding a point E on AC s.t. AB=BE=EC is also helpful. These are two of the “important ideas”, which I were referring to in my previous comment. Just note that this is nowhere near a full solution, you would still have to do one additional construction and a slick intermediate proof, but it’s a start.

Whether they are actually correct, who knows. I personally couldn’t work it out. Showing the length of AB should be equal to BE and EC’s is not that obvious to me. Triangle BEC is just isosceles triangle if that’s relevant to anyone.

  1. Someone also posted the question on math stack exchange. No answers though. Probably posted by OP. Never mind, the post is the same with what hansioux has written above.
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:rofl: I’m in tears laughing. scrolling through and just keep seeing different people post the same diagram over and over. We got pencil drawings, CAD drawings, photos of a computer screen drawings, screenshots of other drawings not related to the original…we got it all here… except answers

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Someone posted a solution on the stack exchange question. Using Euclidean geometry, how to find $x$? - Mathematics Stack Exchange

answer
answer652×861 78.5 KB

But 2nd line is already a wrong statement, angle EAD is x not 2x.
Anyways you just need to show that angle EBD is 2x for the final calculation. There’s probably a nice way to do it. But for me, it’s easier after you figure out the other angles from the next few steps.

The solution writer fixed it. It supposed to be angle EBD = 2 * angle EAD = 2x. By inscribed angle theorem.

And I’m not sure how to justify the 3rd line.

Asked around in the math discord. Not exactly the way he does it. But by construction, we find point E such that the length of AB is equal to BE’s. Because both are radii of the circle centered in B with radius AB, it’s an isosceles triangle. So both angle BAE and AEB is 2x as shown in the picture.

Angle CEB is just 180 deg - 2x in that case. We know that angle ACB is x from the problem statement. So the unknown angle CBE is x. And the triangle BCE is also an isosceles triangle.

As RickRoll said below, use the exterior angle theorem. Angle CBE + angle ACB (x) = angle AEB (2x). So angle CBE = x.

And if 3rd line is correct, it leads to triangle CEB and CDA to be similar/ have same angles. Doesn’t seem obvious that this statement leads to CE/CD = CB/CA. Since we established that BCE triangle is an isosceles triangle with 2 interior angles of x. We see that both triangle CEB and CDA are “similar”. As in they have the same angles and ratio.

From that we get CD/CA = CE/CB. Same ratio of the sides. Rearranging that we get CE/CD = CB/CA as written on the solution.

From that relation CE/CD = CB/CA and the fact that angle DCA = ACB = x from the problem, we see that triangle DCE and ACB are similar. They have the same angles and ratio.

We can conclude that angle EDC must be 2x, because it must be the same as angle BAC.

For the 5th line of the solution, we use the high school exterior angle theorem. So angle DEA must be the sum of the remote angles EDC (2x) and DCE (x), for a total of 3x.

Since we know that angle DEA is 3x and that angle AEB is 2x, the sum of them angle DEB is 5x.

Both BD and BE are radii of the circle. So triangle BDE is an isosceles triangle, i.e. angle BDE must be the same as angle DEB = 5x we found just now.

Angle EBD can be determined several ways. Using the fact that the sum of interior angles of a triangle is 180 degrees, the high school exterior angle theorem, etc. There’s probably a smart way to do it. But dumb way worked for me. Anyways, it’s 2x. Found using the inscribed angle theorem.

Summing all of the angles inside triangle BDE → 2x + 5x + 5x = 12x = 180 degrees, we get x = 15 degrees.

One of the comments on the question also suggested a solution using sine rule. Getting the equation sin 6x sin 2x = (sin 3x)^2 and show it only applies to x = 15 degrees. Dunno about the detailed steps though.

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You can use the same theorem to justify the 3rd line, no?

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You are correct.

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With " the sum of interior angles" way, all I get is ∠DBE=180-10x.
I don’t know how to use the “high school exterior angle theorem” here.
Care to share the “dumb way”? Maybe it’s not that dumb afterwards :wink:

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I mean just brute force it like what we did before the solution on stack exchange. We calculated that the the angles in the crossing of AC and BD lines are either 4x or 180 - 4x.

We know that angle AEB is 2x. So if the wider angle is 180 - 4x, the other angle EBD must be 2x too.

For the small triangle, say triangle BEO, where O is the point where the lines AC and BD intersect each other.

The solution writer on math stack exchange fixed it. He used inscribed angle theorem to show that angle EBD is 2x.

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Oh, I see it now!
Thanks!

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Well, I think we got our answer.
Is that good enough for you, @Chemist116 ?

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I have been asked elsewhere to end the debate. As moderator, this is a simple request to fulfill to keep our punters happy!

Thank you all for playing.

Now I have to hop over to another most productive thread where people are challenged to do algebra to solve for x without doing any algebras.

:lock:

:angry:

11173 113

:face_with_peeking_eye:

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@urodacus
Thanks a lot. Although there is a small lockpad next to the thread. Is it opened right?. I am asking this due to it may confuse someone in the main forum if wants to stop by and reply here but may not do so due the small icon.

Actually not. The solution posted by @overflow does uses triangle similarity and if you look at my original post I clearly mentioned that I was constraining the solution to avoid that route. In other words to no use it.

The thing is that this problem belongs to a chapter which has only introduced the following concepts:

  • The sum of angles in a triangle equals 180 degrees because of the identity from the sum of internal angles in a polygon. S=180(n-2), where n is the number of vertex in this case n=3.

  • Triangle congruence, cases SAS, SSS and ASA.
    Special lines in triangles and their intersections, orthocenter, barycenter, incenter, circumcenter and excenter.

  • As a consequence from the above, it is likely that using circles may be allowed as they appear as a result of the radius formed either by the circumcenter or incenter. For obvious reasons, with more or less difficulty the alluded high school identity of the inscribed angle can be derived easily.

  • Angles formed by the intersection of angle bisectors in a triangle. That is the intersection of the internal angles bisectors of two vertices which makes up for x=90+a/2. The intersection of two external angle bisectors which is x=90-a/2 and the intersection of an internal angle bisector with an external angle bisector which is x=a/2.

  • Special right triangles, with those being 30-60-90, 45-45-90, 37-53-90 and 15-75-90.

  • Identity that relates that the line connecting the midpoints of a triangle is half of the size of its base.

Thus using only these and no theorems relating similarity it should be solved.

There is a solution which uses similarity as mentioned and is relatively simple than the construction made but again this isn’t exactly the answer that I was looking for thus I asked if somebody known a way to do it bypassing this advanced concept. And I say this in terms of my book point of view of building up skills.

Now looking at the proposed solution it is somewhat difficult to catch the part regarding on how angle EDC is the same as angle EAB.

The geometrical argument discussed uses similarity case SAS (side angle side). Assuming that the angle in two triangles is congruent but their sides are proportional to a known ratio.

Then there is that because triangle CEB is similar to triangle CDA from the criteria of angle-angle (AA) with those being angle DAE same as DCE same as ECB and angle EBC. Thus goes on to say from this a ratio is CE/CB=CD/CA
, I changed to this notation in the answer is stated as CE/CD=CB/CA, but to me the former is more logical and easier to understand.

Thus that is the ratio, let’s say k.

In order to find angle EDC we need to use similiarity of two triangles once again. Since we have the ratio now all is left is to find a congruent angle to use the criteria of side angle side.

Now comes the coup of grace, that angle DAC = angle DCE = x it establishes the congruent angle between triangle ACB and triangle DCE. Which had been chosen on purpose to relate their lenghts.

This part is needed to call for the SAS criteria of triangle similarity. We have ratio and the congruent angle.

This part is missing in the original answer,
because CE=kCB, this k=CD/CA and CD=kCA.
Therefore looking at those two triangles we establish that the opposing angle to CE is the same as the opposing angle to CB. Therefore angle EDC=2x.

The question here is how did you get to know use triangles DCE with triangle ACB. This part I believe it is ingenuity because there are too many triangles there, but the one which we do know something other than our first two choices is triangle ACB.

At this point you may understand that for every specifical problem in geometry you need some dose of enlightenment to invent a particular solution for such problem. I know that not everyone has this gift. But I believe it isn’t galaxies away to get use to it.

Regarding the rest is just obvious to use the fact that triangle DEB is isosceles and the sum of interior angles in a triangle adds up to 180 we get 15 which is our answer.

However this solution is not the allowed method of solving. As mentioned lines above is out of the scope from the topics presented so far in my book and as such it would be some sort of maybe early bird or puzzle presented as challenge. Upon inspecting the order of chapters as similarity is presented way after, I still assume that the author intended that the student would solve it with the already known tools.

So all and all, does someone has other ideas?. I mentioned above that I may want to try contact Mathematical socienty in Taiwan for this. I haven’t writen yet but maybe they will be interested. I’d appreciate this remains open so it may gather some potential players or whoever known to solve this by the allowed methods.

By the way @11173 are you also into geometry?.

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no

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your bullet 2 congruence cases utilize similarity. What specific step in the solution do you think is not covered by the topics you mentioned?

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Please note that similarity is not congruence. It may cause some confusion because the acronyms for each cases share the same letters.

Congruence in simple terms means two identical figures because they share the same lenghts and angles. Similarity takes into account enlarged or reduced figures, hence ratios. More on this is explained here and also here.

Similarity and congruence use similar strategy on the essense that they require some construction but the way how this is done is different.

As for your question, I will say again, that all the answer is acceptable until the part that uses similarity two times AA and SAS.
Notice triangles CDA with CEB for the first, and triangles DCE with ACB.

In both cases they use ratios they don’t say they’re equal. Had this been the case then it would had been easier to establish those cases, while for jumping on to SAS you need to do AA to spot the ratio.

That’s why I mentioned the criteria of a congruent angle but two proportional sides.

I shouldn’t do this because as I said, these concepts are introduced later in my textbook because the figures and all the rest has greater difficulty.

Since this problem is presented in an earlier chapter I am assuming that the author had a way to do it without needing similarity. Therefore as the requested solution has not yet been solved.

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guess I misspoke when I said utilizes similarity. it is similar. I didn’t recall the solution utilizing the similar, non congruent triangles; thanks.for pointing out the specifics.