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$AB$ 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.

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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 anglesIllustration of points and circle

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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 $\;\angle AOM=\angle BOM\;$ ...

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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.

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