## Sunday, April 29, 2012

### Problem 745: Circle, 90 Degrees, Chord, Congruence, Perpendicular

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

Click the figure below to see the problem 745 details.

3. http://img715.imageshack.us/img715/7306/problem745.png

Since ∠ACD=90 => AD is a diameter
Since AB=BD => BO perpendicular to AD
Let BE cut AD at F and r = radius of the circle
Since EF//CD => AE/AC=AF/AD => AF=10/7 .r
Note that ∠EAF=∠OBF= alpha
In ∆ BOF , tan(alpha)=OF/OB=3/7
So x=3

4. B is the midpoint of arc AD and BE is perpendicular to chord AC. Hence by the Broken Chord Theorem, 5 = 2 + x or x = 3
If you wish to prove the theorem all you need to do is extend AC to F such that CF =CD=x. Now /_ABD = /_ACD = 2*/_AFD and B being the midpoint of arc AD lies on the perpendicular bisector of AD. Therefore B is the circumcentre of triangle ADF and BE is perpendicular to AF so AE = EF or 5 = 2 + x or x = 3
Ajit

5. http://img18.imageshack.us/img18/6838/p745resuelto.png

(look at the picture)

Note that ABCD is cyclic, then 7*BD=BD*x+2sqrt(2)*BD*sqrt(2)
<=>x=3

Greetings

6. AD is a diameter so < BDA = 45 = < BCA. Hence BE = EC = 2 and so applying Pythagoras to Tr. ABE, AB^2 = 29 from which we have AD^2 = 58.
So x^2 = 58 - 7^2 = 9 and thus x = 3

Sumith Peiris
Moratuwa
Sri Lanka

7. Another proof without using Pythagoras...

Extend EC to F such that CF = 3. So Tr. ABF is isoceles and hence B is the centre of circle ADF. Hence < AFD = 1/2 of < ABD = 45 and so x = CF = 3

Sumith Peiris
Moratuwa
Sri Lanka

8. Still another method much simpler than both previous proofs...,

Find G on AE such that GE = 2. Then prove that Tr.s AGB & BDC are congruent SAA and the result follows.

Sumith Peiris
Moratuwa
Sri Lanka

9. Problem 745
From the theorem of broken chord (Archimedes) is arc AB=arc BD then AE=EC+CD.
So 5=2+x .Therefore x=3
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
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