Circle Geometry..FUN (1 Viewer)

[z600]

Yepp I ran into trouble while doing these, any help would be great!

thanks in advance

Questions ii) iii) and iv)

Question iv) I know you suppose to prove that the angle is half the other..but how

 

[chousta]

hints:

1.ii. use scale ratios of triangles found earlier in (i) result comes from that

iii. use part (ii) and look for a common angle! and answer is a simple statement regarding those 2 facts (ie what can you say about 2 corresponding sides and their included angle )

iv. use alternate segment theorm
then note the cresponding similarity in trianlges
where you can make use of the parallel lines

as a result proving base anlges of an isocles triangle.


if noone gets the next question ill post up answers, or if u need extra help just holla ill do full solutions.

 

[ssglain]

Use Chousta's hints for Q1.
Q2: Think about the isoceles triangles you've proven in parts ii) & iii). Remember equal arcs subtend equal angles at circumference.

Full solutions for Q1 & Q2 are below:

Spoiler

Q1.
ii) TA/TF = TE/TA (Sides of similar triangles are proportional)
.: TA² = TE.TF

But, TA = TB (Tangents from external point are equal)

Hence TB² = TE.TF

iii) This proof is exactly the same to i)

iv) ABT = ADB (Alternate segment theorem)
i.e. FBT = EDB

Now, BET = FBT (by similarity)
But BET = EBD (Alternate angles of parallel lines are equal)
i.e. FBT = EBD

.: EDB = EBD
Hence triangle DEB is isoceles (Base angles equal)

Q2.

iv) From previous parts, triangles AMH and AHL are isoceles.
MAF = FAH & HAE = EAL (Altitude of isoceles triangle bisects vertex)
i.e. MAB = BAK & KAC = CAL
.: arc(MB) = arc(BK) & arc(KC) = arc(CL) (Equal arcs subtend equal angles at circumference)

Now, arc(BKC) = arc(BK) + arc(KC)

.: arc(MKL)
= arc(MB) + arc(BK) + arc(KC) + arc(CL)
= 2[arc(BK) + arc(KC)]
= 2arc(BKC)

Hence arc(BKC) = (1/2)*arc(MKL)

 

[victorheaven]

For the first question
part ii, TFA similar to TAE
TF/TA=TA/TE ==>TE*TF=TA*TA (1)
But TA and TB are tangents from an external point
therefore TA=TB, sub into (1) and part ii is done

part iii, use the result from ii
TE*TF=TB*TB ==> TE/TB=TB/TF
and <FTB is the common angle
therefore they are similar

part iv
<TBF=<ADB (angle between tangent and ...) (2)
<ABD= <ABE+<EBD= <AFE
<AFE=<TFB ==> <EBF+<EBD= <TFB (3)
Also , since EBT similar to BFT
<TFB=<EBT= <TBF+ <EBF (4)
combine (3) and (4) u get,
<EBD= <TBF
therefore <EBD= <ADB (from (2))=>isoceles

 

[victorheaven]

lol.. second one is done alrdy..
gl z600

 

[chousta]

actually ssglain , and victorheaven your proofs for question 1 part 3 wouldnt get you marks. thats not sufficient enough just to say that....and its not "EXACTLY THE SAME AS PART I"



Angle BTE= Angle FTB (common)
(TE/TB)=(TB/TF), from part (ii)

therefore, triangle EBT similar to triangle BFT ( 2 pairs of corresponding sies are in proportion and their including angles are equal), QED

 

[ssglain]

actually ssglain , and victorheaven your proofs for question 1 part 3 wouldnt get you marks. thats not sufficient enough just to say that....and its not "EXACTLY THE SAME AS PART I"

I did not write a proof for pt iii). I was simply telling z600 to do pt iii) in the same manner as s/he did pt i). Which is also a FULLY VALID METHOD - seeing as you enjoy the use of capital letters for dramatic effect.

Nonetheless, I agree your method is easier. Thank you for clarifying this for the others.

 

[chousta]

not a problem only joking around...on another note what school do you go to? and what was your predicated estimate for UNSW interview. ps congrats on that.

 

[ssglain]

not a problem only joking around...on another note what school do you go to? and what was your predicated estimate for UNSW interview. ps congrats on that.

Thanks! Did you also apply? I hope you got an offer, too.

I suspect that if we continue to chat about non-maths things, these posts will be deleted, if the mods are in a foul mood. So: MATHS MATHS MATHS MATHS MATHS and MORE MATHS.

I go to Burwood GHS. I was given 98-100. How about you? I'm still a bit worried that my UMAT won't be too good though, but I'll see next Thursday.

 

[z600]

yepp, solved them both before i came back....thanks alot guys

 

[victorheaven]

actually ssglain , and victorheaven your proofs for question 1 part 3 wouldnt get you marks. thats not sufficient enough just to say that....and its not "EXACTLY THE SAME AS PART I"



Angle BTE= Angle FTB (common)
(TE/TB)=(TB/TF), from part (ii)

therefore, triangle EBT similar to triangle BFT ( 2 pairs of corresponding sies are in proportion and their including angles are equal), QED

LOL... i put this 'and <FTB is the common angle'
but it was not displayed.. dunno y

 

[victorheaven]

actually ssglain , and victorheaven your proofs for question 1 part 3 wouldnt get you marks. thats not sufficient enough just to say that....and its not "EXACTLY THE SAME AS PART I"



Angle BTE= Angle FTB (common)
(TE/TB)=(TB/TF), from part (ii)

therefore, triangle EBT similar to triangle BFT ( 2 pairs of corresponding sies are in proportion and their including angles are equal), QED

LOL... i put this ' FTB is the common angle'
but it was not displayed.. dunno y