# Find the value of $\theta$ in parallelogram $ABCD$.

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!

• Original question image Dec 19, 2020 at 12:04
• If $BD$ is a straight line, then $\triangle ABD \cong \triangle CBD$ and $\theta=40^\circ$. Dec 19, 2020 at 12:08
• Yes, that’s what I said. Dec 19, 2020 at 12:11
• @Tavish How $\triangle ABD \cong \triangle CBD$? Dec 19, 2020 at 12:15
• @Tavish I think by that condition $\triangle ABD \cong \triangle CDB$ , not $\triangle ABD \cong \triangle CBD$. Dec 19, 2020 at 12:24

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 $$\angle CBO = \angle DAP = \angle DOP = \angle CDO = 40°$$.

• Clever and elegant af Dec 19, 2020 at 15:01
• How $AODP$ is cyclic? Dec 19, 2020 at 15:23
• Pick any point U on line DC left of D and any point V on line AB left of A. Let DA and PO intersect at K, then KPA=PAV and UDP=DPK=DAO. $\theta+20^{\circ}+PAV+DAO=180^{\circ}$. Dec 19, 2020 at 15:34
• $\angle PDA = \angle POA$ also implies $AODP$ is cyclic. This is converse of angles in the same segment. Dec 19, 2020 at 17:01
• @Divide1918 Thanks 🙂 Dec 28, 2020 at 13:12

The following solution assumes $$BD$$ is a straight line.

Extend $$AO$$ such that it intersects $$BC$$ at $$X$$, and extend $$CO$$ such that it intersects $$AB$$ at $$Y$$. Then $$\angle AOY+\angle OAB=\angle OYB=\angle OCB+\angle CBA\implies \angle AOY=\angle COX=100^{\circ}-\theta$$. Then $$\angle AXB=120^{\circ}-\theta$$, and angle sum in $$\triangle BOX$$ gives $$\angle BOX=\angle AOD=60^{\circ}$$. So $$\angle COB=160^{\circ}-\theta$$.

Let the height $$EF$$ be as in OP's attempt. Suppose that $$OE:OF=m:n$$. By similar triangles, $$OE:OF=DO:BO=CO:OY=CD:BY=AO:OX=AD:BX=m:n$$, and $$CO:AO=OX:OY=CX:AY$$ since $$\triangle COX \sim \triangle AOY$$. Now, $$CO:OY=AO:OX\implies CO:AO=OY:OX$$, but $$CO:AO=OX:OY$$, so $$OY:OX=OX:OY\implies OX^2=OY^2\implies OX=OY$$. Hence $$AO=CO$$ and $$\triangle BOC\cong\triangle BOA$$.

Thus $$\theta=40^{\circ}$$.