Proposed Problem

Click the figure below to see the complete problem 393 about Triangle, Orthocenter, Circumcircles, Congruence, Collinear.

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Complete Problem 393

Level: High School, SAT Prep, College geometry

## Monday, November 23, 2009

### Problem 393: Triangle, Orthocenter, Circumcircles, Congruence, Collinear

Labels:
altitude,
circumcenter,
circumcircle,
collinear,
congruence,
orthocenter,
triangle

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1. Call O ,the center of circle 1. Call R, radius of circle 1

ReplyDeleteBD cut circle 1 at D’ .

AC is the perpendicular bisector of DD’ (property of orthocenter)

Triangle ADC is congruent to tri. AD’C (case SSS)

So radius of circle ADC= radius of circle AD’C= R

With the same logic, we will get radius of circles 1, 2,3,4,5 congruent.

2. Trapezoid ODD’G is isosceles so DG=OD’=R

Similarly DE=DF= R and D is the center of circle GEF

3. DE , DF and DG cut circles 2, 3 and 4 at H, M, N .

We have DH=DM=DN= 2R and circle 6 center D radius 2R will tangent to circles 2, 3, 4 at H,M,N

4. Angle (HBD)= angle(DBM)=90 so H,B,M is collinear .

similar logic will be applied for other points.

5. H is the intension of DE so D,E,H collinear.

similar logic will be applied for other points.

Peter Tran