How do you find the angle made by two isosceles triangles in a kite?

[SparkyChem]

The problem is as follows:

In the figure. Find x.

According to my book the answer is 15 degrees.

However I don't know how to solve for this problem. Since I am not good with constructions, perhaps someone here can help me?

The thing is that this must be solved using only synthetic euclidean geometry. No trigonometry.

There's more. It is not allowed to use similarity, but congruency and other simple relations in triangles can be used.

It is to be noted, can you please attach a drawing so I could understand the solution?.

With this requirements. Can it be solved?. I know it can be challenging but maybe someone of you guys may know. Please, help?.

 

[BradtheRad]

The easy rule that comes to mind is all 3 angles of a triangle add up to 180 degrees. I believe you can make simultaneous equations from observations of several triangles.

* Total angle B = 180-3x

* Triangle ABD is isosceles (because x+2x=3x). Line AB=BD.
* Unknown angle ABD=180-6x

* Triangle ACD is isosceles. Line AD=DC.

* Unknown angle BDC =180-5x.

* Line CA goes through midpoint of BD (because x+x=total angle C).

I'm not certain but no doubt a few more observations and some deductive reasoning ought to lead to a provable solution.

 

[Aussie Susan]

@BradtheRad - that is exactly how I started to answer this question but then I saw that we are not to use trig. Therefore I'm not sure that this is the approach the OP is looking for.
On the other hand, why make the problem harder than it needs to be!!! Unless this is a homework problem to test the understanding of something taught in class.
Susan

 

[SparkyChem]

The easy rule that comes to mind is all 3 angles of a triangle add up to 180 degrees. I believe you can make simultaneous equations from observations of several triangles.

* Total angle B = 180-3x

* Triangle ABD is isosceles (because x+2x=3x). Line AB=BD.
* Unknown angle ABD=180-6x

* Triangle ACD is isosceles. Line AD=DC.

* Unknown angle BDC =180-5x.

* Line CA goes through midpoint of BD (because x+x=total angle C).

I'm not certain but no doubt a few more observations and some deductive reasoning ought to lead to a provable solution.

Yes, I recall the identity you mention. For any polygon the sum of their interior angles is given by,

S=180(n-2), where n=sides

But this problem is a little bit more complicated. It requires a construction. Otherwise it cannot be solved. If somebody has ideas on how or what to do, it would be of great help.

By the way, I disagree. AC does not passes by the midpoint of BD. Although the angle bisector theorem in a triangle states something like this is bounded by the fact that it must be isosceles and triangle BCD is not isosceles or at least we don't have sufficient information to affirm this.

Still anyone can help here?. I am stuck. How would you solve this without requiring similarity and only congruence or simple constructions?.

@BradtheRad - that is exactly how I started to answer this question but then I saw that we are not to use trig. Therefore I'm not sure that this is the approach the OP is looking for.
On the other hand, why make the problem harder than it needs to be!!! Unless this is a homework problem to test the understanding of something taught in class.
Susan

Thank you for your valuable feedback, it is just that this problem belongs to my geometry textbook and I am not sure how to solve it. Surely is challenging, but that's how it is. Sorry about that.

Other than this, what do you suggest?. By the way, when I meant that trigonometry functions are not allowed, it is because this problem is intended to be solved without them.

The intended approach is relying in synthetic geometry or euclidean geometry as it is known more commonly, and that is relying in euclidean constructions like lines, circles and all that stuff. Thus, as I mentioned above, does any of you guys have any idea on what to do?.

 

[andre_luis]

From the above picture you can take 5 equations and 5 variables ( 4 angles and x itself ).
By making proper substitutions you can find x, and the remaining angles as a bonus:

From Δ ABC : 2x + ∡ABD + ∡DBC + x = 180 °
From Δ ADC : x + 3x + ∡BDC + x = 180 °
From Δ BDC : x + ∡BDC + x + ∡DBC = 180 °
From Δ ABD : 2x + x + ∡ABD + 3x = 180 °
From ▢ ABCD : x + 2x + 3x + ∡ABD + ∡BDC + 2x = 360 °

--- Updated Jun 29, 2022 ---

Just reviewed the math and fixed it...
Not needed so much equations:

By simetry, angles ∡CBD=∡DBC, so at the end there are only 3 unknown angles, not 4.
Therefore, in theory, by choosing only 4 of the above equantions suffice.

--- Updated Jun 29, 2022 ---

Fixed terms as pointed out by wwfeldman.

 

[BradtheRad]

By the way, I disagree. AC does not passes by the midpoint of BD. Although the angle bisector theorem in a triangle states something like this is bounded by the fact that it must be isosceles and triangle BCD is not isosceles or at least we don't have sufficient information to affirm this.

You are correct.

trigonometry functions are not allowed, it is because this problem is intended to be solved without them.

I believe it means we cannot use sines/cosines/tangents.
Euclidean tells me it's not spherical geometry. A triangle on a sphere can have its angles adding up to more than 180.

 

[wwfeldman]

To SparkyChem
i checked the math, 15 degrees works.
i tried equations like Andre_Luis, but got nowhere, primarily because the unknown
angles and the unknown x are not independent, so simultaneous equations will not work

i don't see what construction will help, at least not yet
suggest you try dropping a perpendicular from one of the vertices
that will give you two right triangles, which may help

does the figure lie in one plane, or is it a pyramid, not a kite?

to Andre_Luis:
i think triangle BDC equation has a total of 2x, not 3x
also, i think the "square" has a total of 8x not 9x

 

[wwfeldman]

to SparkyChem:
does the question in your geometry book say anything else?
did you leave anything out?

 

[andre_luis]

In addition to wwfeldman's remark, please note that given the number of equations exceeds the number of variables by one, if these equations are really linearly independent, the answer would be parametric, which means that some detail may indeed be missing in the statement, such as if some edges are parallel for example. .

 

[Aussie Susan]

Just actually looked at the title for this question! The overall figure is supposed to be a kite which means that the lines from the apex and base points on the outside have to be equal. Given that the kite looks like it is drawn on its side, its looks like the lines AB and AD should be equal in length and also lines BC and DC should be the other pair of equal length lines.
That also means that the lines AC and BD meet at a right angle in the middle. Lets call that crossing point X.
Looking at the triangle AXD, the angles must add to 180 so 4x+90 = 180 or x=90/4 = 22.5 degrees.
BUT
It also means that triangle ABD must be isosceles because AB and AD are the same length if that is (say) the apex of the kite. However we are told that the angle BAD is 3x and the angle ADB is 3x. That means the triangle ABD is in fact equilateral so the 3rd angle ABD must also be 3x. Therefore 9x=180 or x = 20.
I suspect that the drawing has been badly labelled as I have interpreted the angle BAC as being 2x. However I think this label really apples to the angle BAD (appropriately named in this case!!!) Therefore the angles BAD and BCD are equal which means the outer shape is a rhombus (which is a subset of a kite).
BUT
If angle BAC is really x then the triangle can't be isosceles but we are told that it is.

TL;DR - My feeling is that there is something wrong with the figure as provided and/or the information in the title.
Of course I could be interpreting the whole thing incorrectly with my tired old brain!
Susan

 

[wwfeldman]

@AussieSusan
i think you analyzed the problem well

i did similar things 3 different ways and landed in a contradiction,
or that the unknowns are not independent (since the are interrelated angles of triangles, they aren't independent)

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