Theorems of the circumference (geometry)

Question

I was studying the circumference and theorems of that in my book. There are two theorems that I wanted to demonstrate, but, when I did it, it somehow disappointed me and I´d like to know how you´d prove it, if you need me to give my demonstration it´s ok, but I don´t, I think It could demonstrated in other ways but don´t know how. I´ll show them in pictures.

I´d also like to know how to prove the theorem that says, that if I have a tangent line, that touches the circumference in $P$, is perpendicular to radius drawn from the centre that point $P$.

Thanks in advance.

Answer 1

See the attached figure.

The first result can be proved from the second result. In particular, if $\angle BAC =180^o$ then $\angle BDC =90^o$. The proof of the second result is very simple.

In the figure, $\triangle ABD$ is isosceles. Therefore, $$\angle ABD = \angle ADB = \alpha.$$Also, $\triangle ACD$ is isosceles. Consequently, $$\angle ACD = \angle ADC = \beta.$$ Further, $$\angle BAD = 180^\circ - 2\alpha,$$and $$\angle BAE = 180^\circ - \angle BAD = 2\alpha.$$ Likewise, $$\angle EAC = 2 \beta.$$ Therefore, $$\angle BAC = 2 \alpha + 2\beta = 2(\alpha + \beta)=2\angle BDC.$$

Regarding the tangent line result, I give you a hint. A line segment drawn from a point (center of the circle in your case) to a line (the tangent line in your case) has the shortest length when the segment is perpendicular to the line.

Answer 2

For a direct proof of the first problem (without using the second result), let $E$ be the other intersection of $DA$ with the circle. Then $\,BECD\,$ is a parallelogram since the diagonals bisect $\,AD=AE=R$, $AB=AC=R\,$ where $\,R\,$ is the radius of the circle, and is in fact a rectangle since $\,BC=DE=2R\,$ and a parallelogram whose diagonals are equal is a rectangle.

(To prove that a parallelogram with equal diagonals is a rectangle, if not obvious, show for example that $\,\triangle DBE \equiv \triangle BDC\,$ as having all sides pairwise equal, which implies $\,\angle DBE = \angle BDC\,$, but on the other hand $\,\angle DBE + \angle BDC = 180^\circ\,$ as consecutive interior angles in a parallelogram.)