# proof of perpendicular lines in a circle

$AB$ is a chord of a circle, centre $O$, and $M$ is its midpoint . The radius from $O$ is drawn through the midpoint $M$. Prove that $OM$ is perpendicular to $AB$.

I know that the product of perpendicular lines is $(-1)$ but i dont know how to express this problem as a proof.

Lets focus on triangles MOA and MOB

• OA = OB because O is center of circle and A and B are on the circle

• AM = BM since M is midpoint of AB

• Triangles MOA and MOB have MO in common

• Therefore triangles MOA and MOB have same sides and same angles but are mirrored along OM.

• This implies Angle OMA = Angle OMB .

• Also these two angles are supplementary since OM falls on AB

Equal supplementary angles are right angles: Angles OMA and OMB are right angles Hints:

Look at the triangle $\;\Delta AOB\;$ and observe this is an isosceles triangle (why?) with basis $\;AB\;$.

But then $\;OM\;$ is the median to the basis in an isosceles triangle, thus...

BTW, you also have that $\;\angle AOM=\angle BOM\;$ ...

Notice $OA = OB = r$, where $r$ is the radius of circle. Since $AM = MB$ by hypothesis, $\Delta AMO \cong \Delta BMO$. Also, we have $\angle AOM = \angle BOM = \gamma$. Similarly, $\angle MAO = \angle MBO = \beta$. We want to show that $\angle BMO = \alpha = 90^o$. Well, since sum of angles in a triangle is $180$, then

$$2\gamma + 2 \beta = 180 \implies \gamma + \beta = 90$$

Since $\alpha + \gamma + \alpha = 180$, then

$\alpha = 90$ as desired.