Is this geometry problem uniquely solvable?

Question

Knowing AC = BC, find x.

I tried to solve the question and found many angles.

And x + k = 110, x + y = 140, k + z = 130, z + y = 160.

But I still can't find x. Is this problem uniquely solvable?

The book answer is x = 30°.

See https://www.duckware.com/tech/worldshardesteasygeometryproblem.html — u1571372, Nov 09 '16 at 21:47
You do have almost finish because calling $O$ the point $DB\cap AE$ you have dans le quadrilatere $OECD$ the sum $$130^{\circ}+90^{\circ}+110^{\circ}+x=360^{\circ}$$ Hence $x=30^{\circ}$ — Ataulfo, Nov 10 '16 at 00:59

Seyed · Accepted Answer · 2016-11-10T01:07:26.690

6

Yes this problem is solvable and there are distinct solutions.

And this is pure geometric solution:

And more easier solution is:

And my final solution:

edited Nov 10 '16 at 01:07

answered Nov 09 '16 at 23:41

Seyed

You mean distinct way to solve because the solution is unique – Ataulfo Nov 10 '16 at 00:40
1

@Seyed: Since the question has been closed as a duplicate, you might want to move your answer to this previous question, where it may get more attention over time. Also, you shouldn't use images as main content of your post. Doing so makes your text impossible to search and inaccessible to the visually impaired. Please transcribe text and mathematics (note that you can use LaTeX). – Blue Nov 10 '16 at 01:24
1

@Blue, Thanks for your useful advices. I sent my answers to that question. In the future I will try to separate images from the main content. Thanks again. – Seyed Nov 10 '16 at 01:58
Nice compilation of solutions. I like the third one! (+1) – Hypergeometricx Nov 10 '16 at 18:38

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