Can someone please help me with this geometry problem?

Hello. This labeled diagram can represent many different triangles. In other words, there are many possibilities for the sum of x+y. :cool:

Thank you so much it makes sense. What confuses my mind is my teacher sent me this and I couldn't understand how we found the angle ''y'' and why the BDE triangle is a isosceles triangle solved.jpg
 
I cannot read most of your second image. I can see points D and E. I can see that CD and CE have equal length.

Triangle CDE is isosceles. Why do you think triangle BDE is isosceles?
 
I cannot read most of your second image. I can see points D and E. I can see that CD and CE have equal length.

Triangle CDE is isosceles. Why do you think triangle BDE is isosceles?

My teacher told me that. I asked him why and he didn't replied. Thanks for your help I think he made a couple of mistakes when solving it.
 
My teacher told me that [triangle BDE is isosceles] …
Then we should think about this some more; maybe I made an error in logic. I'll try picking some random values for x, to approximate the rest, and see what happens.

In the meantime, can you post a better diagram of your work or explain what you tried?
 
Then we should think about this some more; maybe I made an error in logic. I'll try picking some random values for x, to approximate the rest, and see what happens.

In the meantime, can you post a better diagram of your work or explain what you tried?

this is a better picture thumbnail_IMG_7417.jpg
 
Thanks for posting that. I'll have time to study it later today. Can you also post all of the information given to you, by the teacher?

Thats all the info I've been given. As I've mentioned he also said that BDE is isosceles but I cant see any evidence of that, and he still didn't replied my message so I cant ask him how. Thanky you so much for your help.
 
Thats all the info I've been given. As I've mentioned he also said that BDE is isosceles but I cant see any evidence of that …
I couldn't progress with his hints, and the only triangle that I could come up with through guess-and-check (using Law of Sines and Law of Cosines) shows that BDE is not isosceles.

There's one more thing I'd like to try tomorrow, but here's the triangle I'm working with for now (with AD set to 100 arbitrarily):

AD = AB = BC = 100

AC = 128.5575

BD = 68.4040

CD = 34.7296

BAD = 40°

DAC = DBC = 10°

BCD = 20°

ABD = BDA = 70°

ADC = 140°
 
Are we keeping in mind that it asks for x+y and NOT for either x or y?
 
Are we keeping in mind that it asks for x+y and NOT for either x or y?
I am. ;)

I was hoping to find an equation like this:

x + object + y = 180°

where I could show that object = 90°

This exercise has ruffled my feathers!
 
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With the Center Point labeled "O",
With angle BOA labeled "a",
With angle CAO labeled "b"
With angle AOC lebeled "c"
With angle CBO labeled "d"

We have
x = 10
y = 20
a = 140
b = 80
c = 70
d = 20
x + y = 30

We have
x = 17
y = 3
a = 143
b = 83
c = 67
d = 13
x + y = 20

We have
x = 1
y = 1
a = 177
b = 117
c = 33
d = 29
x + y = 2

It's not a rigid structure.

What am I missing?
 
First of all sorry for my English.I found something but I dont know If I did it right. If you draw a circle around D,P,A corners and if the origin is the C corner then DC=PC=EC=ED and DPEC is a deltoid. We know the DCP is a isosceles triangle so PDC angle = 75 and that means X is 15 degrees. That means BDC and BDE is isosceles.

I draw this using geogebra
GEO1.jpg
And then I write down all the angles and I saw that ABE and BDE triangles are isosceles as well so ABDE is also a deltoid and BMA angle is 90 degrees. And AM is a bisector so AMD angle equals to MAE angle. That means ''y'' angle is 75 degrees. Am I right?
thumbnail_IMG_7441.jpg
 
If you draw a circle around D,P,A corners and if the origin is the C corner then DC=PC=EC=ED and DPEC is a deltoid. We know the DCP is a isosceles triangle so PDC angle = 75 and that means X is 15 degrees. That means BDC and BDE is isosceles … Am I right?
Well, I also got x+y=90°, but my ΔBDE is not isosceles. :cool:

tkhunny posted some other versions.

I suspect that your result is another correct possibility.

Having recently returned from a few hours at my favorite watering hole, I should probably not try to comprehend it all right now. But, it seems like we're back to my original hunch: there's more than one possibility.

I think the exercise needs to provide another piece of information, in order to determine a single shape. For example, if it were given that ΔABD is isosceles, then I think we could easily show that x+y=90°.
 
With the Center Point labeled "O",
With angle BOA labeled "a",
With angle CAO labeled "b"
With angle AOC lebeled "c"
With angle CBO labeled "d"

We have
x = 17
y = 3
a = 143
b = 83
c = 67
d = 13
x + y = 20

We have
x = 1
y = 1
a = 177
b = 117
c = 33
d = 29
x + y = 2
I verified the other case, but I am unable to find a set of six sides that work for either of these two cases.

Can you state some sides?


In the verified case, I got (rounded):

AB = AC = 65.2704

BC = 100

AD = BD = 34.7296

CD = 68.4040
 
… that means X is 15 degrees. That means BDC and BDE is isosceles.

I draw this using geogebra

View attachment 8465

… means ''y'' angle is 75 degrees. Am I right?
Yes, I verified that the following exists (rounded).

AD = AB = BC = 100

AC = 141.4213562

BD = CD = 51.76380904

BAD = 30°

DAC = DBC = BCD = 15°

ABD = BDA = 75°

ADC = 135°

Good work!

Let us know what your teacher says, after you show him that more than one case exists. :cool:
 
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