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• February 9th 2011, 12:40 PM
razemsoft21
if angles DAC , BAC
are equal and DC = CB
show that ABCD is a

Attachment 20735
• February 9th 2011, 05:19 PM
rtblue
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: $\displaystyle \frac{AD}{DC}=\frac{AB}{BC}$.

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

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)
• February 9th 2011, 08:21 PM
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 $\angle ADC=\angle ABC=90^\circ$

I cannot find enough information to show this
• February 10th 2011, 12:22 AM
emakarov
Quote:

Originally Posted by rtblue
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: $\displaystyle \frac{AD}{DC}=\frac{AB}{BC}$.

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.

Quote:

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).

The angle x is not located between the two equal sides. For the terminology see the sufficient conditions for triangle congruence.
• February 10th 2011, 02:29 AM
razemsoft21
Quote:

Originally Posted by rtblue
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: $\displaystyle \frac{AD}{DC}=\frac{AB}{BC}$.

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

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)

This theorem does not work in this situation

Can you prove that AD=AB ?
• February 10th 2011, 04:03 AM
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 $180^o$
• February 10th 2011, 07:00 AM
bjhopper
I see a kite. A kite is cyclic if and only if it hastwo right angles

bjh
• February 10th 2011, 08:02 AM
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.
• February 10th 2011, 01:32 PM
bjhopper
Quote:

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.

A kite is a special quadrilateral and always has an axis of symmetry It is cyclic when formed by two congruent right triangles

bjh
• February 10th 2011, 01:39 PM
Yes, but razemsoft's image is not fully defined in such a way.
• February 10th 2011, 02:47 PM
bjhopper
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
• February 10th 2011, 02:52 PM
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.
• February 10th 2011, 03:12 PM
razemsoft21
Quote:

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.

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
• February 10th 2011, 03:19 PM
As the outer circle's radius is reduced
and it overlaps the inner one, the angle $\omega\rightarrow\ 0$

Then we have a quadrilateral with two $90^o$ edges,
but in the general case of $\omega>0$, the quadrilateral is also cyclic.
• February 10th 2011, 04:44 PM
razemsoft21
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 ?
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