Proposed Problem
Click the figure below to see the complete problem 427 about Triangle, Two Altitudes, Square of a Side.
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Complete Problem 427
Level: High School, SAT Prep, College geometry
Saturday, February 20, 2010
Problem 427: Triangle, Two Altitudes, Square of a Side
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We've AD=bsin(C) & AE=bcos(A). Also AFcos(90-B)=bcos(A). So AF=bcos(A)/sin(B). Hence, AD*AF=b^2*sin(C)cos(A)/sin(B). Similarly, CF*CE=b^2*sin(A)cos(C)/sin(B) which gives us: AD*AF+CE*CF=(b^2/sin(B))[sin(C)cos(A)+sin(A)cos(C)]=(b^2/sin(B))[sin(C+A)]=b^2=AC^2 since A+B+C=180 deg.
ReplyDeleteAjit
▲AFG ~ ▲BFD, ▲FGC ~ ▲EFB (right tr,& ang perpen sides) =>
ReplyDeleteAF/BF = FG/FD => BF∙FG = AF∙FD (1)
FC/BF = FG/EF => BF∙FG = FC∙EF (2)
from (1) & (2)
AF∙FD = FC∙EF
AF∙FD + FC∙EF = 2FC∙EF
AF( AD-AF ) + FC( EC - FC ) = 2FC∙EF
AF∙AD - AF² + FC∙EC - FC² = 2FC∙EF
AF∙AD + FC∙EC = AF² + 2FC∙EF + FC²
AF∙AD + FC∙EC = AE² + EF² + 2FC∙EF + FC²
AF∙AD + FC∙EC = AE² + ( EF + FC )²
AF∙AD + FC∙EC = AE² + EC²
AF∙AD + FC∙EC = AC²
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trazas la perpendicular de F al lado AC y determinas dos cuadrilateros inscrisptibles y aplicas el teorema de las tangentes osea ;
ReplyDeleteAF.AD=AH.AC CF.CE=CH.AC
sumamos las dos ecuaciones :
AF.AD + CF.CE = AC (AH + CH ) pero AH + CH = AC
AF.AD + CF.CE = AC.AC --- Lqqd
era el teorema de las secantes plop -_-!! para la proxima reviso mejor antes de publicar
ReplyDeleteNicely done in the Feb 21, 8:48 post above. I cannot read Spanish, but I think H is the foot of the perpendicular from F to AC. Then circumscribing circles about the cyclic quadrilaterals FEAH and FDHC leads to the first two product equations.
ReplyDeleteF is the orthocentre let BH meet AC at H.
ReplyDeleteFrom similar triangles AF. AD = AH. AC and CF. CE = CH. AC
Adding; AF. AD + CF. CE = AH. AC + CH. AC = AC ^2.
Sumith Peiris
Moratuwa
Sri Lanka
It follows that AF. AD + BF. BH + CF. CE = 2 (AB^2 + BC^2 + CA^2)
ReplyDeleteI think 2 will multiply in left hand side.2(AF. AD + BF. BH + CF. CE) =(AB^2 + BC^2 + CA^2)
DeleteIs AD.AF=AB.AE , CE.CF=CD.CB ( E,B,D,F are conciclic )AB^2=AC^2+BC^2-2.CD.BC,
ReplyDeleteBC^2=AB^2+AC^2-2.AE.AB so AC^2=CD.BC+AE.AB=CE.CF+AD.AF.