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# Secant ABC intersects a circle at A and B. Chord $\overline{BD}$ is drawn. Show that $m\angle CBD\ne \dfrac{1}{2}m\widehat{BD}$.

Last updated date: 18th Jun 2024
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Hint: We produce BO, DO, AD. We recall the definition of a central angle and have $m\widehat{BD}=m\angle BOD$. We use the relationship between the inscribed angle by an arc and central angle $m\angle BAD=\dfrac{1}{2}m\angle BOD$. We use the external angle theorem in triangle ABD and use obtained relations to prove the given statement $m\angle CBD\ne m\widehat{BD}$

Complete step-by-step solution
We observe the given figure in the question. We have the straight-line $ABC$ which cuts the circle with center O at points A and B. The chord BD has been drawn. We produce BO. DO and AD.

We know the measure of the chord is equal to the angle it subtends at the center with its two radii which is known as the central angle. Here the chord BD makes the angle $\angle BOD$ with the radii BO and CO at the center O.So we have,
$m\widehat{BD}=m\angle BOD...(1)$

We also know that the inscribed angle by an arc is the angle the endpoints of the arc subtended at any point on the circle excluding the arc as the vertex. Here the arc BD has endpoints B and D. A is a point not on the arc. The arc BD subtends the inscribed angle $\angle BAD$ at A. We know from the theorem that the inscribed angle is half of the central angle. So we have,
$m\angle BAD=\dfrac{1}{2}m\angle BOD...(2)$
We know from the theorem of external angles that the external angle of the triangle is equal to the sum of two remote interior angles. We observe the triangle ABD where the external angle is $\angle CBD$ and its remote internal angles $\angle BAD,\angle ADB$. So we have
\begin{align} & m\angle CBD=m\angle BAD+m\angle ABD \\ & \Rightarrow m\angle CBD=\dfrac{1}{2}m\angle BOD+m\angle ABD\left( \text{from}\left( 2 \right) \right) \\ & \Rightarrow m\angle CBD=\dfrac{1}{2}m\widehat{BD}+m\angle ABD\left( \text{from}\left( 1 \right) \right) \\ & \Rightarrow m\angle CBD>\dfrac{1}{2}m\widehat{BD} \\ \end{align}
So hence it is proved that $m\angle CBD\ne \dfrac{1}{2}m\widehat{BD}$

Note: We note that the inscribed angle by a semicircle is a right angle. We can use the relation between the central angle and the inscribed angle to prove that the opposite angles of a cyclic quadrilateral are supplementary.