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)
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
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.
The angle x is not located between the two equal sides. For the terminology see the sufficient conditions for triangle congruence.Originally Posted by BAdhi
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.
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.