Circle Geometry Help (1 Viewer)

[Trebla]

Hi. Circle Geometry isn't one my of strengths so now I'm stuck on the following questions:
(1) ABC is a triangle insciribed in a circle. The perperndicular from A onto BC meets it at D and is then produced to meet the circumference at K. The perpendicular from C onto AB meets it at F and is then produced to meet the circumference at G. The two perpendiculars AD and CF meet at the point H.
i) Show that the quadrilaterals AFDC and BFHD are both cyclic
ii) Prove that AB bisects the angle GBH
iii) Prove that GB = BK

(2) A circle with centre A, touches a smaller circle with centre B, externally at a point C. PQ is a direct tangent to the two circles, touching them at points P and Q. The common tangent to both circles passing through C meets PQ at the point D. PA and QB, when produced, meet the circumferences of the two circles at T and R respectively. TR meet the larger circle at S.
i) Show that the points P, C and R are collinear
ii) Show that BD is parallel to the line RCP
ii) SHow that the points P, Q, R, S are concyclic

Any help on them would be greatly appreciated. Thank you.

 

[Riviet]

Hi. Circle Geometry isn't one my of strengths so now I'm stuck on the following questions:
(1) ABC is a triangle insciribed in a circle. The perperndicular from A onto BC meets it at D and is then produced to meet the circumference at K. The perpendicular from C onto AB meets it at F and is then produced to meet the circumference at G. The two perpendiculars AD and CF meet at the point H.
i) Show that the quadrilaterals AFDC and BFHD are both cyclic

I assume you have drawn a diagram yourself, carefully labelling each point correctly.
Now draw FD and let angles be denoted by ^, e.g ^ABC is "angle ABC".
Since AD_|_BC, ^ADC is a right angle. Similarly, ^AFC is also right angled. Looking at interval AC, since ^'s ADC and AFC are standing on AC, then the endpoints of the interval and the vertices of the angles are concyclic.
Hence quadrilateral AFDC is cyclic.
Now in quadrilateral BFHD, ^BFH=90^o (complementary angles on a straight line)
Similarly, ^HDB=90^o.
Observe that ^BFH and ^HDB are opposite angles in the quadrilateral.
Also they add to 180^o (ie they are supplementary).
Remember that if the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.
.'. quadrilateral BFHD is cyclic.
If I get anymore, I'll post up my answers. Hope that helps.

 

[Mountain.Dew]

Hi. Circle Geometry isn't one my of strengths so now I'm stuck on the following questions:
(1) ABC is a triangle insciribed in a circle. The perperndicular from A onto BC meets it at D and is then produced to meet the circumference at K. The perpendicular from C onto AB meets it at F and is then produced to meet the circumference at G. The two perpendiculars AD and CF meet at the point H.
i) Show that the quadrilaterals AFDC and BFHD are both cyclic
ii) Prove that AB bisects the angle GBH
iii) Prove that GB = GK

(2) A circle with centre A, touches a smaller circle with centre B, externally at a point C. PQ is a direct tangent to the two circles, touching them at points P and Q. The common tangent to both circles passing through C meets PQ at the point D. PA and QB, when produced, meet the circumferences of the two circles at T and R respectively. TR meet the larger circle at S.
i) Show that the points P, C and R are collinear
ii) Show that BD is parallel to the line RCP
ii) SHow that the points P, Q, R, S are concyclic

Any help on them would be greatly appreciated. Thank you.

i'll do what i can now

1) (ii)

we need to prove that ^GBF = ^FBH

SO, let us assign a value for ^GBF = x, noting that ^GBF = ^GBA

so, ^ACG = x (angles in same segment, ^GBA = ^ACG)
--> ^FDA = x (angles in same segment of cyclic quad AFDC)
--> ^FBH = ^FDA = x (angles in same segment of cyclic quad BFHD)
--> therefore, ^GBF = ^FBH
--> therefore, AB bisects the angle GBH

thats all 4 now, more answers will come later

 

[Riviet]

iii) Prove that GB = GK

Can you check that those are the exact sides given in the question? Because I think you did a typo. From my diagram angle GBK is definitely reflex but angle GKB is acute, therefore GB=/=GK, since the base angles of an isoscles triangle are equiangular. Having said that, I think you meant "prove GB=BK"?

 

[Trebla]

Edit

Oops....sorry, my bad. You're right, it's GB=BK.
I got the first question now, thanks...
Just need help on the parts of Q2

 

[Trev]

Re: Edit

2. (I forget the proper rules so i'll just try and explain crap - I assume you have drawn what is asked)
i) The radii of the circles are all equal, and the tangents from a common point outside of the circle rule (where they have equal length) make a few 'kite' shapes which you can use angle rules with it etc.
you could also show that ACB is colinear and state that it is similar to PCR if you rotate it by an angle (if you get what I mean...)

Hmm that was pretty crap but eh, maybe it helped... I forget circle geo.

 

[Riviet]

2. i) From the original diagram, construct PCR as well as TQ.
Let ^CRQ=x and ^CQR=y.
Now ^TPQ=^RQP=90 degrees (tangent-radius theorem)
But ^TPQ and ^RQP are co-interior and equal, therefore PT || QR.
Now ^RCQ=^PCT=90 degrees (angle in a semi-circle theorem)
Also ^PTC=^CQR=y (alternate angles on parallel lines equal)
.'. ^TPC=180⁰-^CTP-^PCT
=180⁰-y-90^o
=90-y
Hence ^QPC=90^o-^TPC=90-(90-y)=y
Now by the angle sum of a triangle, ^PCQ=180^o-^QPC-^CQP
=180^o-y-(90-y)=90 degrees
Therefore ^PCR=^PCQ+^RCQ=90+90=180, which is a straight line.
.'. P, C and R are co-linear.

 

[Riviet]

2. ii) Draw BD.
DC=DQ (tangents from an external point theorem)
BR=BQ (radii of circle B)
.'. DQBC is a kite, in which the diagonals interesect at right angles.
Let intersection of DB and CQ be E
Then ^BEC=90 degrees
So ^RCQ+^BEC=90+90=180 degrees
Since ^BEC and ^RCQ are co-interior angles as well as supplementary, then it follows that DB || PCR

 

[Riviet]

Wow, they seem to get shorter and slightly easier.

2.iii) Draw PS.
^PST=90 degrees (angle in a semi-circle theorem)
So ^PSR=180^o-^PST
=180^o-90^o
=90^o
Looking at quadrilateral PQRS, we observe that ^PSR+^PQR= 90^o+90^o=180^o => they are supplementary and opposite angles of the quadrilateral. Hence, PQRS is a cyclic quadrilateral, therefore P, Q, R and S are concyclic.