Theorems Related to Angles in a Circle

Consider an arc $\widehat{AB}$ on a circle with center $O$:

• Central Angle (কেন্দ্রস্থ কোণ): The angle subtended by the arc $\widehat{AB}$ at the center $O$ of the circle, denoted by $ngle AOB$.
• Inscribed Angle / Circumference Angle (বৃত্তস্থ কোণ বা পরিধিস্থ কোণ): The angle subtended by the arc $\widehat{AB}$ at any point $P$ lying on the remaining part of the circle's circumference, denoted by $ngle APB$.

Given (প্রদত্ত): Let $O$ be the center of a circle. Arc $\widehat{APB}$ subtends the central angle $ngle AOB$ at the center, and the inscribed angle $ngle APB$ at point $P$ on the remaining part of the circumference.

To Prove (প্রামাণ্য): $ngle AOB = 2 ngle APB$.

Construction (অঙ্কন): Join point $P$ and center $O$, and extend the line segment $PO$ to a point $D$ outside the center.

Proof (প্রমাণ):

1. In $ riangle AOP$, side $OA = OP$ (radii of the same circle).
   Therefore, $ riangle AOP$ is an isosceles triangle, so base angles are equal:
   $$ngle OAP = ngle OPA$$
2. Side $PO$ of $ riangle AOP$ is extended to point $D$. By the Exterior Angle Theorem of triangles, the exterior angle equals the sum of the two interior opposite angles:
   $$ ext{Ext. } ngle AOD = ngle OAP + ngle OPA$$
   Since $ngle OAP = ngle OPA$, we have:
   $$ngle AOD = ngle OPA + ngle OPA = 2ngle OPA \quad \dots ext{ (Equation 1)}$$
3. Similarly, in $ riangle BOP$, side $OB = OP$ (radii of the same circle).
   Therefore, $ngle OBP = ngle OPB$.
   By the Exterior Angle Theorem:
   $$ ext{Ext. } ngle BOD = ngle OBP + ngle OPB = 2ngle OPB \quad \dots ext{ (Equation 2)}$$
4. Adding Equation (1) and Equation (2):
   $$ngle AOD + ngle BOD = 2ngle OPA + 2ngle OPB = 2(ngle OPA + ngle OPB)$$
   Since $ngle AOD + ngle BOD = ngle AOB$, and $ngle OPA + ngle OPB = ngle APB$, we arrive at:
   $$ngle AOB = 2 ngle APB$$

Proof via Theorem 34: Let $P$ and $Q$ be any two points on the major arc of a circle with center $O$, subtended by chord $AB$. Both $ngle APB$ and $ngle AQB$ lie in the same segment.
By Theorem 34, the central angle $ngle AOB$ is double each inscribed angle subtended by the same arc $\widehat{AB}$:
$$ngle AOB = 2ngle APB \implies ngle APB = rac{1}{2}ngle AOB$$
$$ngle AOB = 2ngle AQB \implies ngle AQB = rac{1}{2}ngle AOB$$
Equating both expressions:
$$ngle APB = ngle AQB$$
Hence, angles in the same segment of a circle are equal. (Q.E.D.)

Proof: If $AB$ is a diameter, arc $AB$ is a semicircle. The central angle $ngle AOB$ is a straight angle ($180^\circ$). By Theorem 34:
$$ngle APB = rac{1}{2} ngle AOB = rac{1}{2} imes 180^\circ = 90^\circ$$
Converse Theorem: If a triangle has a right angle, the circle described with the hypotenuse as diameter passes through the opposite right-angled vertex.

Four or more points are concyclic if a single circle passes through all of them. The converse of Theorem 35 provides a powerful test:

• Orthocenter Altitudes Rider: In $ riangle ABC$, altitudes $AD \perp BC$ and $BE \perp AC$ intersect at $H$. Because $ngle ADB = ngle AEB = 90^\circ$, points $A, B, D, E$ are concyclic. Furthermore, because $ngle HDC + ngle HEC = 90^\circ + 90^\circ = 180^\circ$, quadrilateral $CDHE$ is cyclic.
• Angle Chasing Technique: Always express angles in terms of arc measures or central angles ($ngle APB = rac{1}{2}ngle AOB$). Look for isosceles triangles created by radii ($OA = OB$).
• Reflex Angle Application: For a point $Q$ in the minor segment, $ngle AQB = rac{1}{2}( ext{Reflex } ngle AOB) = rac{1}{2}(360^\circ - ngle AOB) = 180^\circ - rac{1}{2}ngle AOB = 180^\circ - ngle APB$. This directly proves that opposite angles of a cyclic quadrilateral are supplementary!