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.
 A: 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 $\;\angle AOM=\angle BOM\;$ ...
A: 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.
A: 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
A: Step 1: in $∆OAM$ and $∆OBM$,
$AM=BM$ [∴M is the midpoint of AB]
$AO=BO$ [∴Radius of the same circle]
$M=M$   [Common line]
Therefore, $∆OAM≡∆OBM$ [SSS Theorem]
∴ $∠OMA=∠OMB$
Step 2: Since the two angles together make a straight angle and are equal.
Therefore, $OMA=OMB=1$ right angle
$∴ OM⊥AB$
[Proved.]
