if angles DAC , BAC

are equal and DC = CB

show that ABCD is a

cyclic quadrilateral.

Attachment 20735

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- Feb 9th 2011, 12:40 PMrazemsoft21cyclic quadrilateral ...
if angles DAC , BAC

are equal and DC = CB

show that ABCD is a

cyclic quadrilateral.

Attachment 20735 - Feb 9th 2011, 05:19 PMrtblue
Are you familiar with the angle bisector theorm? Well, by this theorm, once we recognize AC as the angle bisector, we can set up a proportion:$\displaystyle \displaystyle \frac{AD}{DC}=\frac{AB}{BC}$.

Since we have DC=BC (let's call these two sides x) we have $\displaystyle \displaystyle \frac{AD}{x}=\frac{AB}{x} \Rightarrow AB=AD$

Now, by definition, we have ABCD as a kite. Look up some properties of a kite, and see if you can go on to prove that it is a cyclic quadrilateral. (Hint: It will be cyclic if it is formed by two right triangles) - Feb 9th 2011, 08:21 PMBAdhi
rtblue, can't we come to your conclusion (i.e. AB=AD) in the following way as well?,

since the two triangles are having an equal angle and two equal sides, all the charasteristics of the two triangles are equal

(I don't know the english word for these triangles, if some one know please let me know).

if this is true,

i think we cannot prove that this is a cyclic quadrilateral. Because if this is a cyclic quadrilateral, due to symmetricity, AC has to be the diammeter of the circle. Which makes $\displaystyle \angle ADC=\angle ABC=90^\circ$

I cannot find enough information to show this - Feb 10th 2011, 12:22 AMemakarov
Shouldn't DCB be a straight line to apply the angle bisector theorem? And AB does not have to be equal to AD: look at this picture.

https://lh4.googleusercontent.com/_S...yclic-quad.png

Quote:

Originally Posted by**BAdhi**

- Feb 10th 2011, 02:29 AMrazemsoft21
- Feb 10th 2011, 04:03 AMArchie Meade
The yellow triangles are congruent in the attached construction.

If you label the angles in these,

you can prove that the opposite angles of the quadrilateral sum to $\displaystyle 180^o$ - Feb 10th 2011, 07:00 AMbjhopper
I see a kite. A kite is cyclic if and only if it hastwo right angles

bjh - Feb 10th 2011, 08:02 AMArchie Meade
The circle for which the concyclic points A, B, C and D are on the circumference

does not need to have it's centre on [AC].

Neither does the quadrilateral have to have an axis of symmetry. - Feb 10th 2011, 01:32 PMbjhoppercyclic quadrilateral
- Feb 10th 2011, 01:39 PMArchie Meade
Yes, but razemsoft's image is not fully defined in such a way.

- Feb 10th 2011, 02:47 PMbjhopper
Hi Archie Meade,

From the OP's givens and drawing proof that this is a kite is relatively easy.If you would like one I will reply

bjh - Feb 10th 2011, 02:52 PMArchie Meade
Why don't you ask razemsoft instead of assuming ?

There is the possibilty that other possible cases are not eliminated.

In the sketch I uploaded, the yellow triangles could have an extremely small vertical height

with the circles practically overlapping.

I've exaggerated it for illustration. - Feb 10th 2011, 03:12 PMrazemsoft21
The question doesn't assume that the shape is a kite.

it assumes 4 pts. ABCD which have the following conditions:

http://www.mathhelpforum.com/math-he...drilateral.jpg - Feb 10th 2011, 03:19 PMArchie Meade
As the outer circle's radius is reduced

and it overlaps the inner one, the angle $\displaystyle \omega\rightarrow\ 0$

Then we have a quadrilateral with two $\displaystyle 90^o$ edges,

but in the general case of $\displaystyle \omega>0$, the quadrilateral is also cyclic. - Feb 10th 2011, 04:44 PMrazemsoft21
Can you prove that:

AC is perpendicular to BD

OR

AC divides BD into 2 equal parts

OR

The angels ACB and ACD are equal ?