Circle Geometry. (2 Viewers)

[Carrotsticks]

Here's a pretty cool circle geometry problem.

From a point A, a tangent is constructed to meet a circle at B. From the point B, a horizontal chord is drawn to meet the circle again at a point G. From an arbitrary point the circle another horizontal chord FE is constructed and extends to meet AB produced at C.

From the point C, another tangent is constructed to meet the circle again at D.

Prove that the the line DG bisects the interval FE.

 

[Sindivyn]

This question makes me sad, does it involve multiple congruent triangle proofs?

 

[Carrotsticks]

Just one actually, but quite tricky.

But there are a couple of ways to do it, as with any other circle geometry problem.

 

[nightweaver066]

lol where did you get this from?

 

[Carrotsticks]

Last problem in the Fort Street Trials.

I had a simplified version (after a bit of working out you can drastically reduce the problem) but decided to post the whole thing.

 

[nightweaver066]

Last problem in the Fort Street Trials.

I had a simplified version (after a bit of working out you can drastically reduce the problem) but decided to post the whole thing.

funnily enough, i think this is the exact same question i did at tutor two weeks ago.

 

[jeffwu95]

funnily enough, i think this is the exact same question i did at tutor two weeks ago.

I did this at tutor 2 weeks ago too lol

 

[Carrotsticks]

Out of curiosity, how did your tutoring place do it?

 

[jeffwu95]

Proved that angle GDB= angle FCB, they proved cyclic quads twice and used perpendicular bisector theorem

 

[Carrotsticks]

Proved that angle GDB= angle FCB, they proved cyclic quads twice and used perpendicular bisector theorem

Care to post it up for others to see?

My method was very similar.

 

[SpiralFlex]

Proved that angle GDB= angle FCB, they proved cyclic quads twice and used perpendicular bisector theorem

You go to Harry's right?

 

[nightweaver066]

So messy..

Basically let < G be θ

Then < DOB = 2θ

We know triangles DOC & BOC are congruent so < DOC is θ

< DFC = θ as FC || GB so corresponding angles

Since < DXC = < DOC, D, X, O and C are concylic points since angles in the same segment are equal

then < OXC = < ODC as angles in the same segment

Since the line from the centre of the circle to the chord is perpendicular, it must bisect it

 

[math man]

i had a similar q i gave to my kids using this diagram, where you had to prove XG= XB

 

[math man]

as an exercise fill in missing reasons, i numbered order i proved each angle

 

[Carrotsticks]

"Proof has been left as an exercise to the reader."

 

[math man]

"Proof has been left as an exercise to the reader."

my favourite

 

[jeffwu95]

woo harrys! cool guy

 

[johnpap]

This is actually a pretty neat consequence of the fact that FDEB is a harmonic quadrilateral
I.e. Since DE/DF = CE/CD = CB/CD = BE/BF, FDEB is a harmonic quadrilateral -
then consider taking a perspective from G of FDEB onto the line FE, F and E stay where they are, D goes to X and B goes to the point S at infinity on the line FE, then since the cross ratio of FDEB is preserved XF/XE = SE/SF = 1, (Because S is at infinity), so XF = XE.

 

[seanieg89]

This is actually a pretty neat consequence of the fact that FDEB is a harmonic quadrilateral
I.e. Since DE/DF = CE/CD = CB/CD = BE/BF, FDEB is a harmonic quadrilateral -
then consider taking a perspective from G of FDEB onto the line FE, F and E stay where they are, D goes to X and B goes to the point S at infinity on the line FE, then since the cross ratio of FDEB is preserved XF/XE = SE/SF = 1, (Because S is at infinity), so XF = XE.

Very clever.

 

[math man]

Last problem in the Fort Street Trials.

I had a simplified version (after a bit of working out you can drastically reduce the problem) but decided to post the whole thing.

i found out that no one from the fort street krew could get this question out lol, and it took du over 20 mins to solve it...i guess there cohort isnt that strong.