## Friday, July 20, 2012

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

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

Click the figure below to see the complete problem 788.

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

2. Join BC, BD, BE and BF.

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.

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

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

Tr.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

5. ...the triangles are similar and the result follows

Sumith Peiris
Moratuwa
Sri Lanka

6. Problem 788
Is <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