# Geometry Problem on triangle, tangents and a circle

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".

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

• Hint: $\angle ADB\cong\angle DEB$
– Blue
Commented Oct 19, 2022 at 23:24
• @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 Commented Oct 19, 2022 at 23:59
• $\angle ADB$ 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.)
– Blue
Commented Oct 20, 2022 at 0:14
• @Blue I already know this theorem. I just do not know how to go from there Commented Oct 20, 2022 at 0:27
• 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.
– Dan
Commented Oct 21, 2022 at 9:13

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.

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).

• nvm about prev comment. it makes sense thank you! Commented Oct 21, 2022 at 2:08
• also shouldn't BDA be coloured the same as DBA (because tangents drawn to a point are equal length and therefore BDA is isosceles)? Commented Oct 21, 2022 at 2:42
• @Meemaw Technically that is correct. I chose the colours to make clear which pairs of angles one would need to prove are equal. Commented Oct 21, 2022 at 6:04

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 $$2t$$.

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

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.