See complete Problem 217 at:

gogeometry.com/problem/p217_right_triangle_altitude_projection_congruence.htm

Level: High School, SAT Prep, College geometry

## Saturday, January 3, 2009

### Elearn Geometry Problem 217: Right triangle, Altitude and Projections

Labels:
altitude,
congruence,
projection,
quadrilateral,
right triangle

Subscribe to:
Post Comments (Atom)

We've by similar triangles, BD = AB * BC/AC and

ReplyDeleteFD = 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

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)

ReplyDeleteSimilarly, 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

ABED is a rectangle and diagonals BD and EF intersect at point I, midpoint of diagonal EF.

ReplyDeleteG, 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

Problem 217

ReplyDeleteLet 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