## Sunday, January 4, 2009

### Elearn Geometry Problem 218: Right triangle, Altitude and Distances

See complete Problem 218 at:
gogeometry.com/problem/p218_right_triangle_altitude_perpendicular.htm

Level: High School, SAT Prep, College geometry

1. DFBE is a rectangle so its diagonals meet at the midpoint O of BD. Then
h=2·OD.
But EHGF is a trapezium (trapezoid) hence OD=1/2(a+b).
Finally h=2.OD=a+b

SG

2. Problem 218
Let FP perpendicular in BD (D,P,B are collinear).Then triangleBFP=triangleHDE(FB=DE,<BFP=<BAC=<EDC ).So BP=EH and FG=PD.
Therefore BD=BP+PD=EH+FG.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

3. We see that the triangles AFD ABC and DEC are similar , therefore we get :

and

b /DC =h/AC
bAC=hDC

adding these to equations and using the fact that AD+DC =AC we get

(a+b)AC=ACh
a+b=h

4. In problem 217 we see that GD=HD :

GD=HD =d

Then from similar triangles DGF and EHD we get :

DH/EH=FG/GD
d/b=a/d
d^2=ab

Now let :

FD=BE= x
FB=DE= y

and apply pythagoras to triangle DGF

x^2=a^2+d^2
x^2=a^2+ab

and apply pythagoras to triangle EHD

y^2=a^2+d^2
y^2=a^2+ab

Then from pythagoras we get the relation :

h^2=x^2+y^2
h^2=a^2+2ab+b^2
h^2=(a+b)^2
h=a+b

5. I made a typo when applying pythagoras to triangle EHD.
It should be

y^2=b^2+ab

not

y^2=a^2+ab