Proposed Problem

Click the figure below to see the complete problem 436 about Tangent Circles, Diameter, Chord, Perpendicular, Congruence, Measurement.

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Complete Problem 436

Level: High School, SAT Prep, College geometry

## Friday, April 23, 2010

### Problem 436: Tangent Circles, Diameter, Chord, Perpendicular, Congruence, Measurement.

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▲AFE ~ ▲AGH => AF/AG = AE/AH (right tr & comm ang A)

ReplyDelete▲AGB ~ ▲AFC => AF/AG = AC/AB (right tr & comm ang A)

=>

AE/AH = AC/AB =>

AB∙AE = AC∙AH =>

(AE + EB)∙AE = (AH+HC)∙AH

AE² + EB∙AE = AH² + HC∙AH

AE² + ED² = AH² + MH² (ED², MH² from euclid Theorem)

AD² = AM²

AD = AM

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Problem 436

ReplyDeleteIs AG^2/AF^2=(AH.AB)/(AE.AC),but AG^2/AF^2=AH^2/AE^2=(AH.AB)/(AE.AC) so AH/AE=AB/AC. But AD^2/AM^2=(AE.AB)/(AH.AC=1.Therefore AD=AM.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

By easy angle chase we get tr. ADG ~ tr. AFD and tr. AMG ~ tr. AFM, from their similitude inferring AD^2=AF.AG=AM^2, done

ReplyDeleteGreetings to u all Gogeometers from Sri Lanka as we try to come to terms with what has happened on Easter Sunday

ReplyDeleteMay God help us

< FCA = < GBA = < AGH hence AM^2 = AF.AG

<ADE = < ABD = < AGD hence AD^2 = AF.AG

So AD = AM

Sumith Peiris

Moratuwa

Sri Lanka

Dear Sumith,

ReplyDeleteMy deepest condolences to you and your country

Thank u Antonio for your concern.

DeleteMy wife and I were on the way to churches 100 km North of Colombo when the news came. Pls pray for us.

Dear Sumith

ReplyDeleteMy condolences to your country. I remember these victims in my prayers

Peter tran ,

California, usa

Thank u very much Peter, really appreciate.

Delete