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# In the given figure, $\Delta {\text{ABC}}$ is an isosceles triangle inscribed in a circle with Centre ${\text{O}}$. If ${\text{AB}} = {\text{AC}}$, prove that ${\text{AP}}$ bisects $\angle {\text{BPC}}$ .

Last updated date: 25th Jun 2024
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Hint: Here, we will use the property of chords which states that equal chords subtend equal angle at the circumference of the circle.
Also, we will use the property of the isosceles triangle which states that any triangle said to be isosceles then its two sides are equal in length.

Step 1: From the below figure, it is given that ${\text{AB}} = {\text{AC}}$:
In $\Delta {\text{ABC}}$ , both of its sides are equal, so we can say that $\Delta {\text{ABC}}$ is an isosceles triangle.
Also, ${\text{AB}}$ and ${\text{AC}}$ are chords of the circle which subtends $\angle {\text{APB}}$ and $\angle {\text{APC}}$ at the circumference of the circle respectively.
$\angle {\text{APB = }}\angle {\text{APC}}$ ($\because$${\text{AB}} = {\text{AC}}$)
So, if $\angle {\text{APB = }}\angle {\text{APC}}$ , then we can say that ${\text{AP}}$ bisects $\angle {\text{BPC}}$.
It is proved that ${\text{AP}}$ bisects $\angle {\text{BPC}}$.
${\text{Chord length}} = 2r\sin \dfrac{\theta }{2}$ , $r$ is the radius of the circle and theta being the angle from the Centre of the circle to the two points of the chord.