Geometry Problem

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

Click the figure below to see the problem 737 details.

## Sunday, March 11, 2012

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

Labels:
circle,
circumscribed,
cyclic quadrilateral,
diameter,
perpendicular,
secant,
tangent

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

ReplyDeleteAC 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

This is a solution proposed by Doc Muralidharan who is a medical practitioner greatly interested in math particularly geometry:

ReplyDeleteLet 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

Kudos to Tran, Ajit and Dr. Murali!

ReplyDeletehttp://s1092.photobucket.com/upload/albums/vprasad_nalluri/

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

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

ReplyDeletePravin, 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)

Yes. Your remark about(vi)is correct.

ReplyDeleteH" is the point where the ⊥ bisector of AD and ray HN intersect.

(or H" is the symmetric of H w.r.t.N)

http://i1092.photobucket.com/albums/i418/vprasad_nalluri/RevisedfigureforGG737.jpg

ReplyDeleteAnother Proof:

ReplyDeleteBE/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,

Hence ∠BEF + ∠EAD = 90° and EF⊥AD

Proof(not sure whether it's correct of not):

ReplyDeleteConstruct FK⊥AD, BL⊥AD

∴BL//FK

connect BD

∵AD is the diameter

∴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

So EF⊥AD

Extend CF to H such that FC = FB = FH. Let AC, BD meet at G.

ReplyDeleteConsidering the angles of cyclic quadrilateral BECG we can see that < BFC = 2.< BEC which is also = to 2<BHC since Tr. BFH is isoceles. So F is the centre of circle BHC which circle also passes thro E and G. So E, F, G is a diameter of this circle the points thus being collinear.

Now G is the orthocentre of Tr. AED, hence EDG extended will be perpendicular to AD

Sumith Peiris

Moratuwa

Sri Lanka