Need help with Circle Geometry (1 Viewer)

[blackops23]

Hi guys, there's two questions regarding the proving of a cyclic quadrilateral that I can't get my head around.

Q1. http://img560.imageshack.us/img560/1328/circlegeometry.jpg

Q2. http://img199.imageshack.us/img199/5615/circlegeometry2.jpg

For this one I tried showing angleBNR = angleFPR, now angleFPR=angleFDR], so I made an attempt to show angleBNR = angleFDR], which evidently failed.

Help would be greatly appreciated.

Thanks guys.

EDIT: HOLY SHIT I GOT Q2, so please just Q1 thanks

 

[Deep Blue]

I tried both and couldn't get them but I'll keep trying on Question 1. How did you get Q2?
Is that all the information you are given for both questions?

 

[Alkanes]

question 1 is fucked... LOL

 

[Deep Blue]

It seems to me as if not enough information is given in each question.

 

[Alkanes]

True.. Very hard to see the picture.. Only proof i can see atm that has something to do with it is like angles in the same segment, that is it..

 

[blackops23]

nope this is the right amount of information.

Q2: I tried to prove angleBNR = angleFPR (exterior angle of cyclic = interior opposite angle)
Ok so:
1. angleFPR = angleFDR (equal angles subtended by FR)
2. Now angleFRD = 90 degrees (as FD is a DIAMETER)

Now look at the angle sum of triangle FDR:
angleFDR + angleDFR + angle FRD = 180
angleFDR + angleDFR + 90 = 180
angleFDR = 90 - angleDFR

Now look at angle sum of triangle ONF:
angleFNO + angleDFR + angleFON = 180
angleFNO + angleDFR + 90 = 180
angleFNO = 90 - angleDFR

THEREFORE angleFNO=angleFDR

but angleFNO=angleBNR (equal vertically opposite angles)

THEREFORE angleBNR=angleFDR --> therefore MNRP is cyclic quadrilateral (exterior angle of quadrilateral = interior opposite angle)





Now Q1, I have no idea... Help would be tremendously appreciated. =)

 

[deterministic]

Let angle (PRA)=y
angle (PBA)=angle(PRA)=y (angle in same segment)
let angle(OSR)=x
angle(ABQ)=angle(OSR)=x (ext. angle of cyclic quad)
angle(PBQ)=x+y (obvious from diagram)
angle(ROS)+x+y=180 (angle sum of triangle)

thus angle(PBQ)+angle(ROS)=180
and hence OBPQ cyclic

 

[blackops23]

Let angle (PRA)=y
angle (PBA)=angle(PRA)=y (angle in same segment)
let angle(OSR)=x
angle(ABQ)=angle(OSR)=x (ext. angle of cyclic quad)
angle(PBQ)=x+y (obvious from diagram)
angle(ROS)+x+y=180 (angle sum of triangle)

thus angle(PBQ)+angle(ROS)=180
and hence OBPQ cyclic

wow wow wow thankyou so much mate!!!

 

[Drongoski]

Do you still need soln to Q2 or has it been answered? If not I can post soln.

 

[Drongoski]

Q2. Can be done in various ways. Here's one.

Join PQ & let angle BFR = @

.: angle BPR = @ (you know why)

AF = BF and angle AFB = 90 (semi-circle angle) ==> angle FAB = angle FBA = 45

.: angle FPB = angle FAB = 45

Now angle FNA = angle BFR + angle FBA = @ + 45

angle FPR = ang FPB + ang BPR = @ + 45

.: re quadrilateral MNRP ext angle FNA = interior opp angle FPR

.: MNRP is cyclic


QED

 

[Drongoski]

EDIT: HOLY SHIT I GOT Q2, so please just Q1 thanks

Oh shit - didn't see this earlier.