G1. Let \(ABCDE\) be a convex pentagon such that \(BC=DE\). Suppose there exists a point \(T\) inside the pentagon satisfying \(TB=TD, TC=TE, \angleTBA=\angleAET\).

Let \(P\) and \(Q\) be the intersections of \(CD\) and \(CT\) with \(AB\), respectively. Let \(R\) and \(S\) be the intersections of \(CD\) and \(DT\) with \(AE\), respectively.

Suppose that \(P, B, A, Q\) and \(R, E, A, S\) are collinear in that order. Prove that \(P, S, Q, R\) are concyclic.

Proof. (bmathnguyen)

First, let’s exploit the given equalities in the problem: \(TB=TD, TE=TC, BC=ED\).\

We see that \(\triangle TDE = \triangle TBC\) (by c.c.c).

From (*) we deduce \(\angle ETD=\angle BTC \Rightarrow \triangle ETC=\triangle BTD \Rightarrow \angle ECV=\angle UDB\) (as both triangles \(\triangle ETC\) and \(\triangle BTD\) are isosceles at \(T\)).

From the problem statement: \(\angle TEA=\angle TBA\), to combine with \(\angle TEC=\angle TBD\) and \(\angle ECT=\angle TDB\) as just shown, we introduce points \(U\) and \(V\).

Let \(U\) be the intersection of \(DT\) and \(AB\), and \(V\) be the intersection of \(CT\) and \(AE\).

Then, \(\triangle CEV \sim \triangle DBU\) (by g.g) with \(\angle E=\angle B\) and \(\angle C=\angle D\).

From (**) we obtain \(\angle QVA=\angle SUA\), thus quadrilateral \(UVQS\) is cyclic.\

Now, to prove that quadrilateral \(PRQS\) is cyclic, it is enough to show that \(UV\) is parallel to \(PR\). This is the moment to apply Thales’ theorem.\

We will prove:

\[\frac{PU}{PA} = \frac{RV}{RA}\]

We have

\[VT = \frac{PU}{PB} \cdot \frac{PB}{PA} = \frac{DU}{DB} \cdot \frac{\sin CDT}{\sin CDB} \cdot \frac{CB}{CA} \cdot \frac{\sin DCB}{\sin DCA}\]

We also have

\[VP = \frac{RV}{RE} \cdot \frac{RE}{RA} = \frac{CV}{CE} \cdot \frac{\sin DCK}{\sin DCE} \cdot \frac{DE}{DA} \cdot \frac{\sin CDE}{\sin ADC}\]

We just need to prove:

\[\frac{DU}{DB} \cdot \frac{CB}{CA} \cdot \frac{CE}{CV} \cdot \frac{DA}{DE} = \frac{\sin DCT}{\sin CDT} \cdot \frac{\sin CDB}{\sin DCB} \cdot \frac{\sin DCA}{\sin ADC} \cdot \frac{\sin CDE}{\sin DCE}\]

This is equivalent to:

\[\frac{DU}{DB} \cdot \frac{CB}{CA} \cdot \frac{CE}{CV} \cdot \frac{DA}{DE} = \frac{TD}{TC} \cdot \frac{BC}{BD} \cdot \frac{DA}{AC} \cdot \frac{DA}{AC} \cdot \frac{EC}{ED}\]

This simplifies to:

\[\frac{DU}{CV} = \frac{TD}{TC}\]

The last equality holds true because from (**), and the fact that triangles \(\triangle TEC\) and \(\triangle TDB\) are isosceles, we have:

\[\frac{DU}{CV} = \frac{DB}{CE} = \frac{TD}{TC}\]