1. ## circles

Hello everyone !
please give me an answer to the given circle geometry questions as fast as possible and if possible please today.

Q1. Triangle ABC is inscribed in a circle. the bisector of angle BAC meets line segment BC at P and the circle at Q. line segment QC is joined. If angle BCQ = 38 degree , what is angle BAC ?

Q2. AbCD is a cyclic trapezium in which AD is parallel to BC. Show that angle B = angle C ?

2. Hello, Aveek Agrawal!

1) Triangle $\displaystyle ABC$ is inscribed in a circle.
The bisector of $\displaystyle \angle BAC$ meets side $\displaystyle BC$ at $\displaystyle P$ and the circle at $\displaystyle Q.$
Line segment QC is drawn.
If $\displaystyle \angle BCQ = 38^o$, find $\displaystyle \angle BAC.$
Code:
                B
* * *
*    / \    *
*     /   \     *
*     /     \     * Q
*     /       \     *
*    /         \ *  *
*   /        *  \   *
*  / θ   *       \  *
* /  * θ          \ *
A *- - - - - - - - -* C
*               *
*           *
* * *
$\displaystyle AQ$ bisects $\displaystyle \angle A\!:\;\;\angle BAQ = \angle QAC = \theta$

Draw chord $\displaystyle QC.$

An inscribed angle is measure by one-half its intercepted arc.

Since $\displaystyle \angle BCQ = 38^o$, then $\displaystyle \text{arc}\,BQ = 76^o$

Since $\displaystyle \angle BAQ \;\;^m_=\;\;\tfrac{1}{2}\text{arc}\,BQ,\;\;\angle BAQ = 38^o$ . . . then $\displaystyle \angle QAC = 38^o$

Therefore: .$\displaystyle \angle BAC \:=\:76^o$

2) $\displaystyle ABCD$ is a cyclic trapezium in which $\displaystyle AD \parallel BC.$
Show that: .$\displaystyle \angle B \:=\: \angle C$
Code:
         A    * * *    D
* - - - - - *
*/             \*
*/               \*
/                 \
B * - - - - - - - - - * C
*                   *
*                   *

*                 *
*               *
*           *
* * *
Since $\displaystyle AD \parallel BC$, then: /$\displaystyle \text{arc}\,AB \,=\,\text{arc}\,DC$
. .
Parallel chords intercept equal arcs.

Add $\displaystyle \text{arc}\,AD$ to both sides:
. . $\displaystyle \text{arc}\,AB + \text{arc}\,AD \;=\;\text{arc}\,DC + \text{arc}\,AD \quad\Rightarrow\quad \text{arc}\,BAD \:=\:\text{arc}\,ADC$

Since: .$\displaystyle \begin{array}{ccc}\angle B & ^m_= & \tfrac{1}{2}\text{arc}\,ADC \\ \\[-4mm] \angle C & ^m_= & \tfrac{1}{2}\text{arc}\,BAD \end{array}\quad \text{then: }\angle B \:=\:\angle C$