Question:
If $ABCD$ is a parallelogram then find the angle $\theta$ in degrees.
What I tried:
I assumed $BD$ a straight line as I don't think this question can be solved if $BD$ is not a straight line.(I will be happy if someone solves this without this assumption)
Construction: Drew $EF$ perpendicular to ($CD$ and $AB$) and $GH$ to ($BC$ and $AD$).
$$\angle DAO=\angle DCO=y$$ Alternate interior angles:$$\angle BDA=\theta\\\angle ABD=40^\circ$$
By Angle Sum Property:
$$\angle DOG=90^\circ-\theta\\\angle GOA=90^\circ-y\\\angle AOF=70^\circ\\\angle FOB=50^\circ\\\angle BOH=90^\circ-\theta\\\angle HOC=70^\circ\\\angle COE=90^\circ-y\\\angle EOD=50^\circ$$
Adding all these and equating to $360^\circ$:
$$240^\circ+360^\circ-2\theta-2y=360^\circ$$
$$\theta+y=120^\circ$$
After this, I drew parallel lines to ($AD$ and $BC$) and ($AB$ and $CD$) through $O$ but didn't get the value of $\theta$.
How to solve this question?
Can this be solved?
Thanks!
Best Answer
The assumption is not necessary. Construct $P$ on the parallel to $AB$ by $O$ such that $PAD\cong OBC$. Then $\angle PDA = 20°$, but $\angle POA = \angle OAB = 20 °$, so $AODP$ is cyclic. But then $\angle CBO = \angle DAP = \angle DOP = \angle CDO = 40°$.