Saturday, February 20, 2010

Problem 427: Triangle, Two Altitudes, Square of a Side

Proposed Problem
Click the figure below to see the complete problem 427 about Triangle, Two Altitudes, Square of a Side.

Complete Problem 427

Level: High School, SAT Prep, College geometry

1. 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.
Ajit

2. ▲AFG ~ ▲BFD, ▲FGC ~ ▲EFB (right tr,& ang perpen sides) =>

AF/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

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3. trazas la perpendicular de F al lado AC y determinas dos cuadrilateros inscrisptibles y aplicas el teorema de las tangentes osea ;
sumamos las dos ecuaciones :
AF.AD + CF.CE = AC (AH + CH ) pero AH + CH = AC
AF.AD + CF.CE = AC.AC --- Lqqd

4. era el teorema de las secantes plop -_-!! para la proxima reviso mejor antes de publicar

5. Nicely 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.

6. F is the orthocentre let BH meet AC at H.

From 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

7. It follows that AF. AD + BF. BH + CF. CE = 2 (AB^2 + BC^2 + CA^2)

8. Is AD.AF=AB.AE , CE.CF=CD.CB ( E,B,D,F are conciclic )AB^2=AC^2+BC^2-2.CD.BC,