Question
$ABCD$ is a cyclic qudrilateral in which: $\text{BC}\parallel\text{AD},\ \angle\text{ADC}=110^\circ$ and $\angle\text{BAC}=50^\circ.$ Find $\angle\text{DAC}.$
✓
Answer
Since, $ABCD$ is a cyclic quadrilateral.
Then, $\angle\text{ABC}+\angle\text{ADC}=180^\circ$
$\Rightarrow\angle\text{ABC}+110^\circ=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-110^\circ=70^\circ$
Since, $AD || BC$ Then, $\angle\text{DAB}+\angle\text{ABC}=180^\circ$ [Co-interior angles]
$\Rightarrow\angle\text{DAC}+50^\circ+70^\circ=180^\circ$
$\Rightarrow\angle\text{DAC}=180^\circ-50^\circ-70^\circ=60^\circ$
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$ii.$ If $BL = BM$, prove that $AB = BC$.
Hint:
$i.\text{AB}=\text{BC}$
$\Rightarrow\frac{1}{2}\text{AB}=\frac{1}{2}\text{BC}$
$\Rightarrow\text{AL}=\text{MC}.$
$ii.\text{BL}=\text{BM}$
$\Rightarrow2\text{BL}=2\text{BM}$
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