Cyclic quadrilateral

**reshma** · February 5th 2009, 10:38 PM

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

**red_dog** · February 6th 2009, 07:57 AM

$\widehat{ABC}=\widehat{ACB}$

$\widehat{ABP}=\widehat{PBC}=\widehat{PAC}=\frac{1} {2}\widehat{ACB}$

$\widehat{ACB}=\widehat{PAC}+\widehat{AQC}=\frac{1} {2}\widehat{ACB}+\widehat{AQC}\Rightarrow$

$\Rightarrow\widehat{AQC}=\frac{1}{2}\widehat{ACB}= \widehat{PAC}\Rightarrow\Delta ACQ$ isosceles $\Rightarrow AC=CQ$

**Soroban** · February 6th 2009, 09:36 AM

Hello, reshma!

1. $ABCP$ are points on the circle.
$ABC$ is an 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: $AC=QC.$

Since $\angle ABC = \angle ACB,\;\text{arc}(AC) = \text{arc}(AB)$

Since $\angle ABP = \angle PBC,\;\text{arc}(AP) = \text{arc}(PC)$

Let $\text{arc}(AP) \,=\,\text{arc}(PC) \,=\,a$
. . Then: . $\text{arc}(AB) \,=\,2a$

$\angle PAC$ is an inscribed angle.
. . It is measured by: . $\tfrac{1}{2}\text{arc}(PC) \,=\,\tfrac{1}{2}a$

$\angle AQB$ is formed by two secants.
. . It is measured by: $\tfrac{1}{2}\left[\text{arc}(AB) - \text{arc}(PC)\right] \;=\;\tfrac{1}{2}\left[2a-a\right] \;=\;\tfrac{1}{2}a$

Hence: . $\angle PAC \,=\,\angle AQB$ , and $\Delta ACQ$ is isosceles.

Therefore: . $AC \,=\,CQ$

**reshma** · February 7th 2009, 11:40 AM

Hi,

Can u pls explain me the below part:

is formed by two secants
It is measured by:

Thanks in advance
Reshma

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