## Saturday, January 3, 2009

### Elearn 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

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

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

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

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

5. We have :

BD =h
FD = BE = m
FB = DE = n

From similarity of triangles DGF an BFD :

GD/FD=FB/BD
a/m=n/h
(1.) ah=mn

From similarity of triangles BHE an DEB :

DH/DE=BE/BD
b/n=m/h
(2.) bh=mn

From equation 1 and 2 we have

ah =bh
a=b