3

Let $\overline{AC}$ and $\overline{BD}$ both be chords of the same circle. Let $\overline{AC}$ and $\overline{BD}$ intersect at $E$.

Then why is $\overline{AE}\cdot \overline{EC}=\overline{DE}\cdot \overline{EB}$ ?

J.R.
  • 17,904
user3058477
  • 73
  • Hi new user! $$\color{red}{\Large\text{Welcome to Math.SE!}}$$ Don't worry about it now (since you're new) but you might like to know that we prefer to use MathJax here :) – Shaun Jan 24 '14 at 14:36

1 Answers1

3

I got one simple proof but I don't think it's correct..

Join $A$ with $B$ and $C$ with $D$,

Then we have: $\angle AEB \cong \angle DEC$
$\angle BAE \cong \angle$CDE

By Triangles with two equal angles are similar we have $\triangle AEB\sim \triangle DEC$.

Thus:

$\frac{AE}{EB} =\frac{DE}{EC}$

Thus $AE\times EC = DE\times EB$

Harish Chandra Rajpoot
  • 37,450
user3058477
  • 73
  • Its really simple and absolutely correct, but I mind its originality though .. ;) – Shivanshu Jan 24 '14 at 14:36
  • It's a proposition from Book III of Euclid's Elements. Compare with Proof 2 here of the Intersecting Chord Theorem. – hardmath Jan 24 '14 at 14:39
  • hey @hardmath thanx and really gr8 site you have given me.., – user3058477 Jan 24 '14 at 14:42
  • Another name of this result is called "power of a point". You maybe interested to know that the truth-ness of the result does not depend on whether the point of intersection (E) is inside or outside the circle. – Mick Feb 11 '14 at 14:16