Saturday, January 3, 2009

Elearn Geometry Problem 217: Right triangle, Altitude and Projections

 Geometry Problem 217: Right triangle, Altitude and Projections.

See complete Problem 217 at:
gogeometry.com/problem/p217_right_triangle_altitude_projection_congruence.htm

Level: High School, SAT Prep, College geometry

4 comments:

  1. We've by similar triangles, BD = AB * BC/AC and
    FD = AB * BC/AC * AB/AC while GD = AB * BC/AC * AB/AC * BC/AC = (AB*BC)^2/AC^3 ------(1)
    Similarly it may be shown that DE=BD*BC/AC from where DH=DE*AB/AC=(AB*BC)^2/BC^3 ----(2)
    By (1)& (2) we've, GD=DH or a=b as depicted in the diagram.
    Ajit: ajitathle@gmail.com

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  2. I feel there are some typos in the proof above and therefore it is incorrect. Here's another way to do it: Tr. FGD and Tr. ABC are similar. Hence,GD/FD=BC/AC or GD=FD*BC/AC. But FD=BE so GD=BE*BC/AC. Now from Tr. BDC we've BE*BC = BD^2. Therefore, GD=BD^2/AC ......(1)
    Similarly, DH/DE=AB/AC or DH = DE*AB/AC = BF*AB/AC=BD^2/AC .......(2)
    Now from (1) & (2) we can say that GD = DH or a=b
    Ajit

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  3. ABED is a rectangle and diagonals BD and EF intersect at point I, midpoint of diagonal EF.
    G, D ,H are the projections of F, I, E over line AC.
    Since I is the midpoint of EF ,so D is the midpoint of GH.

    Peter Tran

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  4. Problem 217
    Let FP perpendicular in BD (D,P,B are collinear).Then triangleBFP=triangleHDE(FB=DE,<BFP=<BAC=<EDC ).So DH=FP=GD.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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