Let AB be a diameter in a given circle and let C be a point on the circle such that OC $\perp$ AB. Let D be an arbitrary point on the small arc AC. Let E be the point of intersection between OC and BD. Let F be the point of intersection of the tangent lines drawn at A and D

enter image description here Prove that:

OBEF is a parallelogram

OEDF is a cyclic trapezoid

What I have so far...

to prove OBEF is a parallelogram I have to show wither opposite sides parallel or opposite sides congruent. I am stuck on how to chase the angles

Any help would be appreciated!

  • $\begingroup$ $OBDF$ is not cyclic, is there a typo? $\endgroup$ – Lazy Lee May 20 '17 at 1:59
  • $\begingroup$ yes there is I am sorry, OEDF $\endgroup$ – Parley May 20 '17 at 1:59

Notice that from $OB = AO$, $\angle OBD = \frac{1}{2}\angle AOD = \angle AOF$ and $\angle OAF = \angle BOE=90^\circ$, we know that $\triangle AOF \cong \triangle OBE$ . Further, notice that $\angle OBD = \angle AOF$ tells us $OF\parallel BE$, and congruency gives us $OF=BE$.

Hence, $OBEF$ is a parallelogram because $OF\parallel BE$ and $OF=BE$. Next, because $OBEF$ is a parallelogram, we have $$DE\parallel OF\implies \angle DEF=\angle OFE = \angle AOF =\angle DOF \implies OEDF \ \ \text{is cyclic.}$$

  • $\begingroup$ How do you know OB =EF? $\endgroup$ – Parley May 20 '17 at 2:00
  • $\begingroup$ Hi, I fixed the proof. Thanks! $\endgroup$ – Lazy Lee May 20 '17 at 2:07
  • $\begingroup$ @Lazy Lee: You were supposed to show $OEDF$ is cyclic. $\endgroup$ – quasi May 20 '17 at 2:25
  • $\begingroup$ Yes, it was just a typo in the end. Thanks for noticing it :) $\endgroup$ – Lazy Lee May 20 '17 at 2:31
  • $\begingroup$ I was under the impression that in order for a quad to be proved to be a parallelogram you had to show that BOTH opposite sides parallel or BOTH opposite sides congruent? $\endgroup$ – Parley May 20 '17 at 3:43

$\angle \, EAO = \angle \, EBO = \angle \, EDO$ because triangles $ABE$ and $BDO$ are isosceles with $AE = BE$ and $OB=OD$ respectively. Therefore, $\angle \, EAO = \angle \, EDO$ implies that $ADEO$ is inscribed in a circle $k$. However by tangency, $\angle \, OAF = 90^{\circ} = \angle \, ODF$ so $AFDO$ is inscribed in a circle. But $k$ is the circumcircle of triangle $ADO$ so the pentagon $AFDEO$ is inscribed in $k$. Consequently, $\angle \, OAF + \angle \, OEF = 180^{\circ}$ and because $\angle \, OAF = 90^{\circ}$ so is $\angle \, OEF = 90^{\circ}$. Observe that $AF \perp AB$ and $OE \perp AB$ so $AF \, || \, OE$, and $EF \perp OE$ and $OA \perp OE$. Thus $AFEO$ is a parallelogram inscribed in circle $k$ so it must be a rectangle. Moreover, $EF \, || \, AO$ and $EF = AO = BO$ and since $O \in AB$, $BO \, || \, EF$ and $BO = EF$ so $BEFO$ is a parallelogram.


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