# Hard Circle geometry question (1 Viewer)

#### Unravel

##### New Member
Using geometry theorems, and not coordinate geometry, prove the following:

Suppose the diagonals of the cyclic quadrilateral ABCD intersect at a point G. Draw perpendiculars from G to sides BC and DA, and call these points of intersection as X and Y respectively. Prove that X and Y are symmetric around the line joining the midpoints of AB and CD

#### Paradoxica

##### -insert title here-
Using geometry theorems, and not coordinate geometry, prove the following:

Suppose the diagonals of the cyclic quadrilateral ABCD intersect at a point G. Draw perpendiculars from G to sides BC and DA, and call these points of intersection as X and Y respectively. Prove that X and Y are symmetric around the line joining the midpoints of AB and CD
Here is an equivalent statement of your problem

Let M be the midpoint of AB and N be the midpoint of CD

Prove that MXNY is a Kite.

#### Unravel

##### New Member
Here is an equivalent statement of your problem

Let M be the midpoint of AB and N be the midpoint of CD

Prove that MXNY is a Kite.
Yeah, cool. So can anyone solve the problem?

#### si2136

##### Well-Known Member
Do you have a diagram which comes with the question?

#### Unravel

##### New Member
Do you have a diagram which comes with the question?
No, is there some detail of the question that's confusing you?

#### Unravel

##### New Member
Surely there must be someone capable enough on BoS to be able to do this question? No?

#### KingOfActing

##### lukewarm mess
Surely there must be someone capable enough on BoS to be able to do this question? No?
I might give it a go tomorrow, I really hate geometry #### Paradoxica

##### -insert title here-
Using geometry theorems, and not coordinate geometry, prove the following:

Suppose the diagonals of the cyclic quadrilateral ABCD intersect at a point G. Draw perpendiculars from G to sides BC and DA, and call these points of intersection as X and Y respectively. Prove that X and Y are symmetric around the line joining the midpoints of AB and CD
Let the midpoints of AB and CD be E and F respectively.

Consider the reflection of G along BC and DA, denoted GX and GY respectively.

By definition, AGDGX and BGCGY form kites.

AEB, DFC, GXGX and GYGY are all collinear.

By Spiral Similarity (or in this special case, Spiral Symmetry), EXFY is also a Kite.

Therefore, X and Y are symmetrical about EF

QED

#### Unravel

##### New Member
Let the midpoints of AB and CD be E and F respectively.

Consider the reflection of G along BC and DA, denoted GX and GY respectively.

By definition, AGDGX and BGCGY form kites.

AEB, DFC, GXGX and GYGY are all collinear.

By Spiral Similarity (or in this special case, Spiral Symmetry), EXFY is also a Kite.

Therefore, X and Y are symmetrical about EF

QED
Paradoxica, you are WRONG!

Spiral similarity does not necessarily imply that EXFY is a kite, or if you meant something else by this statement you need to provide some form of justification.

#### Paradoxica

##### -insert title here-
Paradoxica, you are WRONG!

Spiral similarity does not necessarily imply that EXFY is a kite, or if you meant something else by this statement you need to provide some form of justification.
The two exterior kites are congruent (angle chasing will provide the necessary conditions)

Since the vertices of the inside quadrilateral are all exactly halfway between the corresponding vertices (p=0.5), the quadrilateral formed is similar to the two corresponding quadrilaterals.

Thus a kite.

#### Unravel

##### New Member
The two exterior kites are congruent (angle chasing will provide the necessary conditions)

Since the vertices of the inside quadrilateral are all exactly halfway between the corresponding vertices (p=0.5), the quadrilateral formed is similar to the two corresponding quadrilaterals.

Thus a kite.
Unfortunately for you, the two exterior kites are not congruent, if by congruent you mean that they are identical. For some reason I believe you assumed that the lengths of GX and GY are equal

Also, care to explain the part "Since the vertices of the inside quadrilateral are all exactly halfway between the corresponding vertices (p=0.5), the quadrilateral formed is similar to the two corresponding quadrilaterals."?
You have provided no clear reasoning

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#### Paradoxica

##### -insert title here-
Unfortunately for you, the two exterior kites are not congruent, if by congruent you mean that they are identical. For some reason I believe you assumed that the lengths of GX and GY are equal
ah, my bad, I meant similar.

I'll have a look tomorrow at the situation again (unless someone else does it)

#### Unravel

##### New Member
ah, my bad, I meant similar.

I'll have a look tomorrow at the situation again (unless someone else does it)
Yes please do have a look again tomorrow.

I believe that this problem will remain unsolved for quite some time though, (The solution is actually very tricky)

#### Paradoxica

##### -insert title here-
Yes please do have a look again tomorrow.

I believe that this problem will remain unsolved for quite some time though, (The solution is actually very tricky)
Aha, thank you for verifying my thoughts, this was indeed a challenge, not a cry for help. #### Unravel

##### New Member
Also please explain the part "Since the vertices of the inside quadrilateral are all exactly halfway between the corresponding vertices (p=0.5), the quadrilateral formed is similar to the two corresponding quadrilaterals."

This has no clear justification behind it

#### Unravel

##### New Member
Aha, thank you for verifying my thoughts, this was indeed a challenge, not a cry for help. Yeah, there's a few steps which I thought were very difficult to see, so I'm trying to see whether other people could see (I personally could not crack this problem by myself, had to peek at the solutions)

#### DatAtarLyfe

##### Booty Connoisseur
Someone will get it, dont you worry

#### Paradoxica

##### -insert title here-
Unfortunately for you, the two exterior kites are not congruent, if by congruent you mean that they are identical. For some reason I believe you assumed that the lengths of GX and GY are equal

Also, care to explain the part "Since the vertices of the inside quadrilateral are all exactly halfway between the corresponding vertices (p=0.5), the quadrilateral formed is similar to the two corresponding quadrilaterals."?
You have provided no clear reasoning
Spiral Similarity does imply the quadrilateral we are chasing is similar to the exterior similar kites.

The figures are oriented similarly.

Each of the corresponding vertices are the midpoints of the exterior vertices.

I do not see how this argument is invalid.

There are no assumptions made in this argument.

If you want me to show the exterior kites are similar, I can do that.

#### Unravel

##### New Member
Spiral Similarity does imply the quadrilateral we are chasing is similar to the exterior similar kites.

The figures are oriented similarly.

Each of the corresponding vertices are the midpoints of the exterior vertices.

I do not see how this argument is invalid.

There are no assumptions made in this argument.

If you want me to show the exterior kites are similar, I can do that.

What do you interpret spiral similarity as?