1. ABCP are points on the circle.ABC is a isocelels triangle with AB=AC and angle bisector of B meets circle at P. If AP and BC is extended to meet at Q, show that AC=QC.
Please let me know the steps to solove the above problem
$\displaystyle \widehat{ABC}=\widehat{ACB}$
$\displaystyle \widehat{ABP}=\widehat{PBC}=\widehat{PAC}=\frac{1} {2}\widehat{ACB}$
$\displaystyle \widehat{ACB}=\widehat{PAC}+\widehat{AQC}=\frac{1} {2}\widehat{ACB}+\widehat{AQC}\Rightarrow$
$\displaystyle \Rightarrow\widehat{AQC}=\frac{1}{2}\widehat{ACB}= \widehat{PAC}\Rightarrow\Delta ACQ$ isosceles $\displaystyle \Rightarrow AC=CQ$
Hello, reshma!
1. $\displaystyle ABCP$ are points on the circle.
$\displaystyle ABC$ is an isocelels triangle with $\displaystyle AB=AC$, and angle bisector of $\displaystyle B$ meets circle at $\displaystyle P.$
If $\displaystyle AP$ and $\displaystyle BC$ is extended to meet at $\displaystyle Q$, show that: $\displaystyle AC=QC.$
Since $\displaystyle \angle ABC = \angle ACB,\;\text{arc}(AC) = \text{arc}(AB)$
Since $\displaystyle \angle ABP = \angle PBC,\;\text{arc}(AP) = \text{arc}(PC)$
Let $\displaystyle \text{arc}(AP) \,=\,\text{arc}(PC) \,=\,a$
. . Then: .$\displaystyle \text{arc}(AB) \,=\,2a$
$\displaystyle \angle PAC$ is an inscribed angle.
. . It is measured by: .$\displaystyle \tfrac{1}{2}\text{arc}(PC) \,=\,\tfrac{1}{2}a$
$\displaystyle \angle AQB$ is formed by two secants.
. . It is measured by: $\displaystyle \tfrac{1}{2}\left[\text{arc}(AB) - \text{arc}(PC)\right] \;=\;\tfrac{1}{2}\left[2a-a\right] \;=\;\tfrac{1}{2}a$
Hence: .$\displaystyle \angle PAC \,=\,\angle AQB$, and $\displaystyle \Delta ACQ$ is isosceles.
Therefore: .$\displaystyle AC \,=\,CQ$