# Special quadrangle with 135° and 90°

• Feb 13th 2013, 12:50 PM
jalt
Special quadrangle with 135° and 90°
Hi,
I hope you guys can help me with the following problem:
I have a quadrangle ABCD. <ADC = 135°, <ABC = 90°, [AB] = [BC]
That's all I know, but now I have to prove that [BD] = [AB] = [BC]
I tried some things with equations but it doesn't really work.

Btw: I'm from Austria and I'm not a native speaker, so please feel free to correct my linguistic mistakes. ;)
• Feb 14th 2013, 08:41 AM
johng
Re: Special quadrangle with 135° and 90°
At first I didn't even believe it, but dynamic geometry convinced me otherwise. Here's a solution:

Attachment 27018
• Feb 14th 2013, 09:37 AM
jalt
Re: Special quadrangle with 135° and 90°
Hi,
thanks a lot for your reply and your solution. I know this might sound strange but my teacher told me that there was a solution without any trigonometry.
Unfortunately he isn't gonna tell me until I find it out. :(
I think I tried everything but it seems to be impossible to me. Do you have another idea or can you at least tell me whether it's possible or not?
• Feb 14th 2013, 08:20 PM
MINOANMAN
Re: Special quadrangle with 135° and 90°
Jalt

you do not need all the complicated answers..after all the beuty of Mathematics lies on one word SIMPLICITY.
Anyway the quadrelateral you mention is part of a regular octagone with B as center of the circle... check it.....as simple as such

MINOAS
• Feb 14th 2013, 08:38 PM
johng
Re: Special quadrangle with 135° and 90°
I'm afraid I disagree with the previous answer. A regular octagon has all of its sides equal, but there's no restriction on your quadrilateral that AD = DC.

Back to the problem. Since the result is true, the circumcenter of triangle ACD mus be B; i.e. the intersection of the perpendicular bisectors of AD and DC must be B. I can see no reason why this is true. There's probably a simple geometric argument, but I can't see it.
• Feb 15th 2013, 05:52 AM
MINOANMAN
Re: Special quadrangle with 135° and 90°
John

construct a regular octagone with B as the center of circle.and BA=BD=BC as the radii of the circle .The sides AD=DC .The angle ABC=90 degrees and angle abd=135 degrees.
it is very simple try it.
Minoas
• Feb 15th 2013, 07:54 AM
johng
Re: Special quadrangle with 135° and 90°
Minoanman,

What you say about constructing a regular octagon is certainly true, but it does not answer the original question. The sides AD and DC are not necessarily equal. See the figure in my initial response.
• Feb 15th 2013, 09:18 AM
bjhopper
Re: Special quadrangle with 135° and 90°
construct a right angle at B with equal legs BA and BC.Swing an arc from C to A.Mark a point D anywhere on the arc. Prove that the angle ADC is always 135 deg It is simple
• Feb 15th 2013, 09:33 AM
MINOANMAN
Re: Special quadrangle with 135° and 90°
JOHNG
Construct a regular octagone with B as the center of circle and BA,BC BD RADII . THE QUADRILATERAL ABCD IS ONE OF THE TRILLIONS YOU CAN GET WITH THE SAME PROPERTY ONCE YOU SLIDE THE VERTEX D OVER THE ARC AD. TRY IT IT IS SIMPLE. I AGREE WITH BJHOPPER IT IS SIMPLE.

MINOAS
• Feb 15th 2013, 07:08 PM
johng
Re: Special quadrangle with 135° and 90°
I agree with bjhopper that his construction always produces a quadrilateral as specified originally -- it's nothing more than an application of the inscribed angle theorem. I'm not being obtuse on purpose, but why is every quadrilateral, as originally described, an instance of this construction? It seems to me that you need to know that length BD is the same as BA and BC (of course this is true, but this was the original question).

minoanman: I think my source of confusion with you is that you keep saying regular octagon. As defined everywhere, a regular polygon must have sides of equal length. In particular, a regular octagon has equal sides.
• Feb 16th 2013, 10:32 AM
johng
Re: Special quadrangle with 135° and 90°
Geometry fascinates me, but I've always known my skills at geometry are poor. The missing item in my information bank was:
Given an obtuse triangle ADC with obtuse angle $\theta$ at D, the circumcenter of ADC is the unique point P on the perpendicular bisector of AC which satisfies:
1. P is in that half plane determined by line AC which does not contain D.
2. The angle APC is $2\pi-2\theta$.

So the proof as suggested by bjhopper is absolutely correct. There's nothing magic about 135 degrees, it can be replaced by any $\theta$ and the angle ABC being $2\pi-2\theta$.

MY ignorance caused all the hubbub. Sorry.
• Feb 24th 2013, 03:36 AM
jalt
Re: Special quadrangle with 135° and 90°
Hi,
I am sorry, I was on vacation the last week.
Unfortunately I still haven't understood your solution.

I have an arc with the center B and the radius AB and now I claim that <ADC is always 135° if D is on the arc.
Well that's true but how can I prove it?

@johng What angle is pi?
• Feb 24th 2013, 05:07 AM
bjhopper
Re: Special quadrangle with 135° and 90°
Hi jalt,
See post 8.
Triangle ABD and DBC are isosceles.Sides are radii
Angle BAD=BDA =a Angle BDC =BCD =b Property of Isosceles triangles
360 =90 +135 +2a +2b Property of quads
270 = 2a+2b
a+b = 135 angle D.D can be anywhere on the arc CA
• Feb 24th 2013, 07:04 AM
jalt
Re: Special quadrangle with 135° and 90°
Thanks a lot!
I am still just a little confused, because you proved that D can be anywhere on the arc CA, but you didn't prove that D has to be on the arc.
Why isn't it possible that there is a point outside of the arc that makes <ADC = 135°?
There's probably a very easy solution but I don't see it.
• Feb 24th 2013, 10:57 AM
johng
Re: Special quadrangle with 135° and 90°
Jalt,

I hope the attached diagram and explanation helps:

Attachment 27228