Question

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 \[\angle {\text{BPC}}\] .

Answer 1

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.

Complete step-by-step answer:
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.
Step 2: By using the property of chords that two equal chords subtends an equal angle at the circumference of the circle we get:
 \[\angle {\text{APB = }}\angle {\text{APC}}\] (\[\because \]\[{\text{AB}} = {\text{AC}}\])
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 \[\angle {\text{BPC}}\].

Note: Students should remember that a chord of a circle is a straight-line segment whose endpoints lie on the circle. The longest chord of the circle is its diameter.
The formula for finding the length of the chord of a circle is as shown below:
\[{\text{Chord length}} = 2r\sin \dfrac{\theta }{2}\] , \[r\] is the radius of the circle and theta being the angle from the Centre of the circle to the two points of the chord.