Angles in cyclic quadrilateral in semi-circle

• Sep 26th 2011, 12:08 PM
I-Think
Angles in cyclic quadrilateral in semi-circle
Given a cyclic quadrilateral ABCD in a semi-circle, (let AD be the diameter),let AB=BC=1. Draw the line DB.
Prove that angle ADB= angle BDC

This was taken as obvious in a solution I read, but it is not at all obvious to me. Help in understanding this result would be appreciated.
• Sep 26th 2011, 02:09 PM
Soroban
Re: Angles in cyclic quadrilateral in semi-circle
Hello, I-Think!

Quote:

$\displaystyle \text{Given a cyclic quadrilateral }ABCD\text{ in a semicircle with diamter AD.}$
$\displaystyle \text{Let }AB = BC = 1.\;\text{ Draw the line segment }DB.$
$\displaystyle \text{Prove that: }\:\angle ADB \,=\, \angle BDC$
Didn't they explain why it is "obvious"?

Angles $\displaystyle ADB$ and $\displaystyle BDC$ are inscribed angles.
. . They are measured by one-half their intercepted arcs.

We are told that chords $\displaystyle AB$ and $\displaystyle BC$ are equal.
. . Hence, their respective arcs are equal: .$\displaystyle \overline{AB} \:=\:\overline{BC}$

Therefore: .$\displaystyle \angle ADB \,=\,\angle BDC$