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