Theorems related to Circle

A circle is the plane figure formed by the locus of all points in a two-dimensional plane that maintain a constant distance (radius $r$) from a fixed reference point (center $O$).

Given (প্রদত্ত): Let $O$ be the center of a circle. $AB$ is a chord of the circle which is not a diameter. $M$ is the midpoint of chord $AB$, such that $AM = BM$. Line segment $OM$ joins center $O$ to midpoint $M$.

To Prove (প্রামাণ্য): $OM \perp AB$, which means $ngle OMA = ngle OMB = 90^\circ$ (one right angle).

Construction (অঙ্কন): Join $O, A$ and $O, B$ to form radii $OA$ and $OB$.

Proof (প্রমাণ): In $ riangle OMA$ and $ riangle OMB$:

1. $OA = OB$ (Radii of the same circle).
2. $AM = BM$ (Given, since $M$ is the midpoint of chord $AB$).
3. $OM = OM$ (Common side).

Therefore, by the Side-Side-Side (SSS) congruence criterion:
$$ riangle OMA \cong riangle OMB$$
Since the triangles are congruent, their corresponding angles are equal:
$$ngle OMA = ngle OMB$$

Now, the ray $OM$ stands on the straight line segment $AB$. By Euclid's linear pair axiom, the sum of two adjacent angles is $180^\circ$ (two right angles):
$$ngle OMA + ngle OMB = 180^\circ$$
Since $ngle OMA = ngle OMB$, we substitute:
$$ngle OMA + ngle OMA = 180^\circ \implies 2ngle OMA = 180^\circ \implies ngle OMA = 90^\circ$$
Therefore, $ngle OMA = ngle OMB = 90^\circ$, which proves that $OM \perp AB$. (Q.E.D. / প্রমাণিত)

Given (প্রদত্ত): Let $O$ be the center of a circle and $AB$ be a chord. $OM$ is drawn perpendicular to $AB$ ($OM \perp AB$, so $ngle OMA = ngle OMB = 90^\circ$).

To Prove (প্রামাণ্য): $OM$ bisects chord $AB$, i.e., $AM = BM$ ($M$ is the midpoint of $AB$).

Construction (অঙ্কন): Join $O, A$ and $O, B$ to form radii $OA$ and $OB$.

Proof (প্রমাণ): In right-angled triangles $ riangle OMA$ and $ riangle OMB$:

1. $ngle OMA = ngle OMB = 90^\circ$ (Given, $OM \perp AB$).
2. Hypotenuse $OA = $ Hypotenuse $OB$ (Radii of the same circle).
3. Side $OM$ is common to both triangles.

Therefore, by the Right angle-Hypotenuse-Side (RHS) congruence criterion:
$$ riangle OMA \cong riangle OMB$$
Since corresponding parts of congruent triangles are equal (CPCTC):
$$AM = BM$$
Hence, the perpendicular $OM$ bisects the chord $AB$. (Q.E.D. / প্রমাণিত)

In right-angled triangle $ riangle OMA$, by Pythagoras' theorem:

Let radius $OA = r$, perpendicular distance $OM = d$, and chord length $AB = c$. Since $OM$ bisects $AB$, half-chord $AM = rac{c}{2}$.

• Equidistant Chords Theorem: Equal chords in a circle are equidistant from the center. If $c_1 = c_2$, then $d_1 = d_2$.
• Converse: Chords that are equidistant from the center of a circle are equal in length. If $d_1 = d_2$, then $c_1 = c_2$.
• Inverse Distance Relation: Of two unequal chords in a circle, the longer chord is closer to the center ($c_1 > c_2 \implies d_1 < d_2$). The maximum chord length is attained when $d = 0$, which is the diameter ($c_{ ext{max}} = 2r$).
• Concentric Circles Property: A line intersecting two concentric circles creates equal intercepted segments between outer and inner circumferences ($AB = CD$).