If $ABCD$ is a parallelogram then find the angle $\theta$ in degrees. enter image description here

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 $AD$). Fig 1

$$\angle DAO=\angle DCO=y$$ 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?


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

2 Answers 2


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

enter image description here

  • $\begingroup$ Clever and elegant af $\endgroup$
    – Divide1918
    Dec 19, 2020 at 15:01
  • $\begingroup$ How $AODP$ is cyclic? $\endgroup$
    – Wolgwang
    Dec 19, 2020 at 15:23
  • $\begingroup$ 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. $\theta+20^{\circ}+PAV+DAO=180^{\circ}$. $\endgroup$
    – Divide1918
    Dec 19, 2020 at 15:34
  • $\begingroup$ $\angle PDA = \angle POA$ also implies $AODP$ is cyclic. This is converse of angles in the same segment. $\endgroup$
    – player3236
    Dec 19, 2020 at 17:01
  • $\begingroup$ @Divide1918 Thanks 🙂 $\endgroup$
    – jjagmath
    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 $\theta=40^{\circ}$.


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