## Sunday, March 11, 2012

### Problem 737: Cyclic Quadrilateral, Circumscribed Circle, Diameter, Tangent, Perpendicular

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

Click the figure below to see the problem 737 details.

1. http://img638.imageshack.us/img638/6608/problem737.png

AC cut BD at H and ∠ABD=∠ACD=90 ( see picture)
Quadrilateral BECH is cyclic with EH as a diameter
OF perpendicular to BC => OF is the perpendicular bisector of chord BC
Note that ∠BAC=∠BDC=∠FBC=∠FCB ( angles face the same arc BC)
∠BEC =∠BFK = ½∠ (BFC) ( angles complement to congruence angles ∠FBK and ∠EAC)
On OF only center of circle BECH has property BEC=1/2(BFC)
So F is the center of circle BECH => E, F, H are collinear
H is the orthocenter of triangle AED and EF perpendicular to AD

2. This is a solution proposed by Doc Muralidharan who is a medical practitioner greatly interested in math particularly geometry:
Let AC and BD intersect at H and let EH extended meet AD in Y. Since /_ABD and /_ACD are right angles, BD and AC are the altitudes of the triangle Tr.EAD & H is its orthocenter. EH is therefore perpendicular to AD. Now let F be the midpoint of EH. Join FB and produce it to X
Since Tr.EBH is rt angled at B, F, the midpoint of the hypoteneuseEH is also the circumcentre of the triangle, so FE=FB….......(1)
Since /_EBD=/_EYD=90, E, B, Y, D are concyclic, and so BEF=BDA
But BEF=FBE (from1)=XBA (vertically opp angles). So XBA=BDA
Chord AB subtends equal angles in the segment ADB and the line XBF. By the converse of the alternate angle theorem, line XBF is tangent to the circle at B. In other words, the tangent at B passes through the midpoint of EH.
Repeating the same argument with C, we see that the tangent at C also passes through the midpoint of EH. Therefore the two tangents intersect at F, which is the midpoint of EH, which is part of EY which is perpendicular to AD.
Doc Murali
Goa, India

3. Kudos to Tran, Ajit and Dr. Murali!
It is interesting to note that
(i)the ninepoint circle of ∆ADE passes through O, B, F, C, Y
(ii)F is the midpoint of EH
(iii)O, N, F are collinear
(iv)B, E, C, H are concyclic
(v)E, F, H, Y are collinear
(vi)the reflection H' of H in OF lies on the circle through B, E, C, H
(vii)H' is the circumcentre of ∆ADE

4. http://img204.imageshack.us/img204/7319/notesofproblem737.png

Pravin, see my comments on items (vi) and (vii) below
It is interesting to note that
(i)the ninepoint circle of ∆ADE passes through O, B, F, C, Y [ Yes, ]
(ii)F is the midpoint of EH
(iii)O, N, F are collinear [ Yes, N is the point of intersection of OF and perpendicular bisector of OY]
(iv)B, E, C, H are concyclic [Yes ]
(v)E, F, H, Y are collinear [Yes]
(vi)the reflection H' of H in OF lies on the circle through B, E, C, H [Yes, Since OF pass through center of circle BECH so H’ lies on the circle] see picture)
(vii)H' is the circumcenter of ∆ADE [ No, In my opinion,H' is not the circumcenter of triangle ADE.The circumcenter H” of ∆ADE will be on line HN such that NH”=NH where N is the center of nine point circle] ( see picture)

H" is the point where the ⊥ bisector of AD and ray HN intersect.
(or H" is the symmetric of H w.r.t.N)

7. Another Proof:
BE/BD = tan ∠BDE = tan ∠BDC = tan ∠BCF = tan ∠BOF = BF/BO
So BE/BF = BD/BO. Also ∠EBF = ∠BDA. Thus ∆EBF ~ ∆BOD
∆BOD is isosceles.
∴ so is ∆BEF and ∠BEF = ∠BDO = 90° - ∠EAD,

8. Proof(not sure whether it's correct of not):
∴BL//FK
connect BD
∴BD⊥AE
∴∠BDA=∠AEK
∵△BAD is a right triangle and BL is the altitude of AD
∴∠BDA=∠ABL
∴∠AEK=∠ABL
∴BL//FE (converse of corresponding angles postulate)
∵BL//FE, BL//FK
∴E,F,K have to be collinear