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.

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