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Old 9th August 2012, 06:13 PM   #1 (permalink)
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Find the length

Here is a short problem i created (should only require up to Sec 4 Additional math i guess ) :
ABC is a triangle such that AB=16cm and CB=24cm. Point D is on AB such that DB=6cm. Point E is on CB such that EB=4cm. It is also know that AE=12cm and CD=18cm. Let F be the point where AE and CD intersect. Find the length of FE.

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Old 9th August 2012, 07:38 PM   #2 (permalink)
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Re: Find the length

i didnt see got F, so i not sure how to do, very hard

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Old 9th August 2012, 08:42 PM   #3 (permalink)
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Re: Find the length

Here
I have a gut feeling that there's a simpler solution though...
Since AB:AE:BE = 16: 12:4 = 4:3:1 = 24:18:6 = CB:CD:BD, △ABE~△CBD.
→∠BAE=∠BCD and ∠ADC = 180-∠BDC = 180-∠BEA =∠AEC
→ △ADF~△CEF (AA)
→ FE/2=FD and CF/2=AF

AE+CD=30
FE+AF+CF+FD=30
FE+(FE/2)+CF+(CF/2)=30
1.5(FE+CF)=30
FE+CF=20 —— ①
CF+FD=18 —— ②

①-②:
FE-FD=2
FE/2=2
FE=4cm

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Last edited by Betakuwe; 9th August 2012 at 09:44 PM.
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Old 12th August 2012, 07:08 PM   #4 (permalink)
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Re: Find the length

Originally Posted by Betakuwe View Post
Here
I have a gut feeling that there's a simpler solution though...
Yes that is correct.
Here is my solution:
ABC is similar to EBD (SAS)
Angle CAB= Angle DEB
So, Angle CED= 180 - Angle DEB= 180 - Angle CAB
As, Angle CED +Angle CAB =180, ADEB is a cyclic quad.
Hence ADF is similar to CEF (AA) also by a ratio of 1 to 2
Let FD=x, so FE=2x
Now, by intersecting cords theorem, AF*FE=CF*FD or,
(12 - 2x)(2x)=(18-x)(x)
Solving,we will get x=2 or FE= 2*2=4cm.

Solving by similar triangles all the way would be the easiest approach, but i like circle properties, so i used it...

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