Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1244.

## Saturday, August 13, 2016

### Geometry Problem 1244: Circle, Radius, Perpendicular, Chord, Secant, Measurement. Mind Map

Labels:
chord,
circle,
measurement,
mind map,
perpendicular,
radius,
secant

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https://goo.gl/photos/FGTGW9iSkgxRKfx16

ReplyDeleteConnect BF, BC and BD

Note that C is the midpoint of arc BG

Triangle BEC similar to FBC ( case AA)

So CB^2=CE.CF

Triangle DCB similar to BCA ( case AA)

So CB^2=CD.CA

And CE.CF=CD.CA or 4x 10= 3(x+3)

So x= 31/3

Let AB cut the circle at G.

ReplyDeleteIn isoceles Tr. BCG,

CG^2 - CE^2 = BE.GE = CE.EF = 24

So CG^2 = 24 + 16 = 40

Now < CGB = CBG = < GDA

Hence < CGD = < DGB - < CGB = < DGB - < GDA = < GAD

Hence CG^2 = CD.CA

So 40 = 3(3+x) from whence x = 31/3

Sumith Peiris

Moratuwa

Sri Lanka

2nd solution

ReplyDelete< CFD = < CGD = < EAD as before

Hence AFED is cyclic and the result is easily calculated

Sumith Peiris

Moratuwa

Sri Lanka

Join points F.O.D then WE CAN eaisly solve for r.

ReplyDeleteNote that FO=OD=OC=r so Triangle FCD is a right triangle

By similar angle WE have traingle FCD is similar to ECA so

4/3+x=3/10 then X=31/3

First you will need to prove the collinearity of F.O.D

Delete