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)

ReplyDeleteIs 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.