See complete Problem 218 at:

gogeometry.com/problem/p218_right_triangle_altitude_perpendicular.htm

Level: High School, SAT Prep, College geometry

## Sunday, January 4, 2009

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

Labels:
altitude,
congruence,
perpendicular,
right triangle

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DFBE is a rectangle so its diagonals meet at the midpoint O of BD. Then

ReplyDeleteh=2·OD.

But EHGF is a trapezium (trapezoid) hence OD=1/2(a+b).

Finally h=2.OD=a+b

SG

Problem 218

ReplyDeleteLet 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

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

ReplyDeletea/AD=h/AC

aAC=hAD

and

b /DC =h/AC

bAC=hDC

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

(a+b)AC=(AD+DC)h

(a+b)AC=ACh

a+b=h

In problem 217 we see that GD=HD :

ReplyDeleteGD=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

I made a typo when applying pythagoras to triangle EHD.

ReplyDeleteIt should be

y^2=b^2+ab

not

y^2=a^2+ab