Home > Shapes > Geometry
| Rectangle-Diagonal (Posted on 2016-11-02) |
 |
Let E be the foot of the perpendicular from A to diagonal
BD in rectangle ABCD. Let F and G be points on sides BC
and CD respectively such that EFCG is a rectangle.
If |BD| = d, |EG| = g, and |EF| = 1, find d in terms of g.
|
|
Submitted by Bractals
|
|
No Rating
|
|
|
Solution:
|
(Hide)
|
Let |EB| = b and |EF| = f.
Pythagoras' theorem applied to ΔBFE gives b2 = f2 + 1.
ΔAEB ~ ΔBFE ⇒ |AB|/b = |AB|/|BE| = |BE|/|EF| = b/1
⇒ |AB| = b2.
ΔDAB ~ ΔBFE ⇒ d/b2 = |BD|/|AD| = |BE|/|EF| = b/1
⇒ d = b3 ⇒ b2
= d2/3.
|DG| = |CD| - |CG| = |AB| - |EF| = b2 - 1 = f2.
ΔEGB ~ ΔBFE ⇒ g/f2 = |EG|/|DG| = |BF|/|EF| = f/1
⇒ g = f3 ⇒ f2
= g2/3.
b2 = f2 + 1 ⇒ d2/3 = g2/3 + 1 ⇒ d = ( g2/3 + 1 )
3/2.
QED
|
Comments: (
You must be logged in to post comments.)
| |
Subject |
Author |
Date |
 | Solution | Jer | 2016-11-02 14:26:28 |
|
|
Please log in:
Forums (1)
Newest Problems
Random Problem
FAQ |
About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
|
