Geometry Problem

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

Click the figure below to see the complete problem 788.

## Friday, July 20, 2012

### Problem 788: Intersecting Circles, Secant Line, Midpoint, Cyclic Quadrilateral

Labels:
cyclic quadrilateral,
intersecting circles,
midpoint

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http://img687.imageshack.us/img687/6215/problem788.png

ReplyDeleteDraw lines per attached sketch

We have <BDA=<BFA=.5*<BQA=<BQO

And <BEA=<BCA=.5*<BOA=<BOQ

So all 3 triangles OBQ, EBF and CBD are similar per case AA

Tri. CBD is the image of Tri. EBF in the spiral similarity transformation centered B , angle of rotation = theta=<EBC=<FBD=<FAD

Since G is the image of H in this transformation so <HBG=<FAD= theta

So quadrilateral AHBG is cyclic

Join BC, BD, BE and BF.

ReplyDeleteAngle AFB = angle ADB

Angle AEB = angle ACB

Thus, triangle BCD ~ triangle BEF.

Now rotate triangle BCD to triangle BEF,

with center B, such that

BC and BE coinside, BD and BF coinside,

with different lengths.

Then

CD becomes EF (with different lengths),

BG becomes BH (with different lengths).

Therefore, the rotation angle is

angle GBH, and also same as angle CAE.

Hence, AHBG is concyclic.

Let M be the midpoint of OQ. Join OA, QA, MB

ReplyDeletePower of H w.r.t. circle (O) = HA.HE = OA² - OH²

Power of H w.r.t. circle (Q)= HA.HF = HQ² - QA²

So OA² - OH² = HQ² - QA² (since HE = HF),

OA² + QA² = OH² + HQ² which is same as

2AM² + MQ²/2 = 2HM² + MQ²/2

Follows MA = MH

Similarly equating powers of G w.r.t. circle (Q),it can be shown that MA = MG

OQ being the perpendicular bisector of AB,we have MA = MB

Hence MA = MH = MG = MB and A, H, B, G are concyclic.

Let CG = GD = p and let EH = HF = q

ReplyDeleteTr.s BCD and BFE are similar so,

2p/2q = s/t where BC = s and BE = t

Hence s/p = t/q

So in Tr. s BEH and BCG

< E = < C and the sides which hold these angles are proportional

Hence

...the triangles are similar and the result follows

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

Problem 788

ReplyDeleteIs <ACB=<AEB,<ADB=<AFB then triangleCBD similar triangleEBF.So BD/BF=CD/EF=2GD/2HF=GD/HF.Then triangleBDG similar triangleBFH .

So <BGD=<BHF=<BHA.Therefore AHBG is cyclic.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE