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.
$\endgroup$ 14 Answers
$\begingroup$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 $\;\angle AOM=\angle BOM\;$ ...
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.]
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