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