# circles

• Nov 16th 2008, 02:03 AM
Aveek Agrawal
circles
Hello everyone !
please give me an answer to the given circle geometry questions as fast as possible and if possible please today.(Doh)

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 ?(Thinking)

Q2. AbCD is a cyclic trapezium in which AD is parallel to BC. Show that angle B = angle C ?(Wondering)
• Nov 16th 2008, 07:01 AM
Soroban
Hello, Aveek Agrawal!

Quote:

1) Triangle $ABC$ is inscribed in a circle.
The bisector of $\angle BAC$ meets side $BC$ at $P$ and the circle at $Q.$
Line segment QC is drawn.
If $\angle BCQ = 38^o$, find $\angle BAC.$

Code:

                B               * * *           *    / \    *         *    /  \    *       *    /    \    * Q       *    /      \    *       *    /        \ *  *       *  /        *  \  *       *  / θ  *      \  *       * /  * θ          \ *     A *- - - - - - - - -* C         *              *           *          *               * * *
$AQ$ bisects $\angle A\!:\;\;\angle BAQ = \angle QAC = \theta$

Draw chord $QC.$

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

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

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

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

Quote:

2) $ABCD$ is a cyclic trapezium in which $AD \parallel BC.$
Show that: . $\angle B \:=\: \angle C$

Code:

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

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

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