Prove that ##AE=2BC## -Deductive Geometry

[SammyS]

Kindly find my rejoinder...sorry am a bit busy had to scribble my response.
I have made use of similarity (on the diagram) as is suggested...and I still cannot see how $BC= EB$.

That $BC=EB$ does not come from similarity of triangles. It comes from the fact that chord $BC$ and chord $EB$ each subtend inscribed angles of equal measure, namely $\angle BDC$ and $\angle EDB$ .

Steve beat me to it !.

 

[anuttarasammyak]

The essence of the problem is
AC/CD=AE/BC
coming from same arc lengths and similarity of triangles.

Or $\triangle EOB \equiv \triangle COB$ instead of same arc lengths, if you prefer.

 

[chwala]

...just going through my posts...this was a nice one...just to remember how i understand it in the near future; i reflected $AE$ on the supposedly mirror line $AC$, it then follows that;

$\dfrac{AE}{BE}=\dfrac{AC}{CD}$

$\dfrac{AE}{BC}=\dfrac{AC}{CD}$

We are given; $AC=2CD$

$\dfrac{AE}{BC}=\dfrac{2CD}{CD}$

$\dfrac{AE}{BC}=2$

$⇒AE=2BC$.