circles

Aveek Agrawal · #1 $Old$ 11-16-2008, 04:03 AM

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 ?

Soroban · #2 $Old$ 11-16-2008, 09:01 AM

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