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I have tried to solve this problem for very long all to no avail and none of my peers have been able to find a solution either. Can someone provide any clues?

"In the diagram B and D are points of contact of tangents drawn from A to the circle. E is a point on the circle and C is the point on BE such that AC ∥ DE."

"Prove ABCD is a cyclic quadrilateral".

enter image description here

What I know is that for it to be a cyclic quadrilateral, the opposite angles have to be supplementary and that the angle in the tangent is equal to the angle in the alternate segment. I already know that angleBED = angle BDA (alternate angle theorem) but still cannot make headway. But despite tons of angle labelling, I can't seem to get anything at all!

pls help

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  • $\begingroup$ Hint: $\angle ADB\cong\angle DEB$ $\endgroup$
    – Blue
    Commented Oct 19, 2022 at 23:24
  • $\begingroup$ @Blue I have edited my post to account for my knowing this equivalence already. I also marked the other pair of equal angles in alternate segments, and also, by two tangents, I know CBA and CDA are equal. That's all I've really got lol $\endgroup$ Commented Oct 19, 2022 at 23:59
  • $\begingroup$ $\angle ADB$ and $\angle DEB$ are congruent by the Inscribed Angle Theorem (the first angle being a special case of such a thing). More-or-less a converse of that Theorem will get you to the target result. (The directness of that argument depends upon things you've already learned about circle geometry.) $\endgroup$
    – Blue
    Commented Oct 20, 2022 at 0:14
  • $\begingroup$ @Blue I already know this theorem. I just do not know how to go from there $\endgroup$ Commented Oct 20, 2022 at 0:27
  • $\begingroup$ We can also show that $ABCOD$ is a cyclic pentagon, where $O$ is the centre of the circle. The circle with diameter $OA$ contains $B$ and $D$ (because $\triangle{ABO}$ and $\triangle{ADO}$ are right triangles). So, by virtue of the theorem in this question, it also contains $C$. So $ABCOD$ is a cyclic pentagon. $\endgroup$
    – Dan
    Commented Oct 21, 2022 at 9:13

4 Answers 4

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enter image description here

The red lines are cut by a transversal BCE so we have $\alpha=\beta$ as equal alternate angles.

Angles in the alternate segment at D and E are equal so we have $\beta=\gamma$. Remove common angle $\beta$

So $ \alpha=\gamma$ are seen as equal angles on same side of AB, and by virtue of the converse theorem that when equal angles are subtended on same side of AB at C and D, we should have quadrilateral ABCD concyclic inside the green circle.

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Hint: The converse of "angles in the same segment" holds. That is, if two angles subtended from a single line segment are equal and on the same side of the line segment, then the four points they define form a cyclic quadrilateral.

This reduces the problem to showing any one of the below pairs are equal (some are easier than others).

Image of angles

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  • $\begingroup$ nvm about prev comment. it makes sense thank you! $\endgroup$ Commented Oct 21, 2022 at 2:08
  • $\begingroup$ also shouldn't BDA be coloured the same as DBA (because tangents drawn to a point are equal length and therefore BDA is isosceles)? $\endgroup$ Commented Oct 21, 2022 at 2:42
  • $\begingroup$ @Meemaw Technically that is correct. I chose the colours to make clear which pairs of angles one would need to prove are equal. $\endgroup$ Commented Oct 21, 2022 at 6:04
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In figure $1$, radius $\overline{OB}$ and $\overline{OD}$ determine a cyclic quadrilateral $ABOD$ because they are perpendicular to tangents $\overline{AB}$ and $\overline{AD}$ respectively so we have $\angle{BAD}=180^{\circ}-2t$.

Besides, because of $\overline{ED}$ is parallel to $\overline{CA}$ we have $\angle{BCA}=t$. Consequently, in order that $ABCD$ be a cyclic quadrilateral it is necessary that $\angle{BCA}=\angle{ACD}=t$ because $\angle{BCD}$ should be suplementary of $\angle{BAD}$.

On the other hand the corresponding circumcircle is easily constructed in figure $2$ since it is the circumcircle of the triangle $\triangle{ABO}$ of radius $\dfrac{\overline{OA}}{2}$ and center the midpoint of $\overline{OA}$. This circle shows that in fact $\angle{BCA}=\angle{ACD}$ because $\overline{AB}=\overline{AD}$.

NOTE.- Not only points $A,B,C,D$ are on the same circle but also the center of the original circle of the problem. We could say that the irregular pentagon $ABCOD$ is cyclic and that if the first circle is centered at the origine the equation of the circle solution is $x^2-2rx+y^2=0$ where $2r=\overline{OA}$. Finally, on the red circle, point $C$ distinct of $A$ and $D$ can be any in the arc able to segment $\overline{BD}$ seen under the angle $2t$.

enter image description here

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Since $\angle ECD$ is the supplement of $\angle DCB$, $ABCD$ is a cyclic quadrilateral if$$\angle ECD=\angle BAD$$ from kite to cyclic quad

We start with a kite $HAGJ$, touching its inscribed circle at $B$, $D$, $E$, $F$, and complete the isosceles trapezoid $BDEF$.

Let kite diagonal $AJ$ meet trapezoid diagonal $BE$ at $C$. From the bilateral symmetry of the kite and isosceles trapezoid, it is clear that $C$ lies on the line of symmetry $HG$ and on the other trapezoid diagonal $DF$, and further that $AC\parallel DE$ since both are perpendicular to $HG$.

Therefore, by symmetry $\triangle CED$ is isosceles.

And since$$\angle CED=\angle ABD$$(tangent/alternate segment), then$$\triangle CED\sim\triangle ABD$$and$$\angle ECD=\angle BAD$$Conversely then: Given any triangle $BED$ inscribed in a circle, with tangents $AB$, $AD$, and with $AC$ drawn parallel to base $ED$, it is clear from the symmetry of the constructible kite and isosceles trapezoid that$$\angle ECD=\angle BAD$$ whence $\angle BAD$ is supplementary to $\angle BCD$, and $ABCD$ is a cyclic quadrilateral.

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