The diagram shows triangle ABC together with its inscribed circle. Show that [AD], [BE] and [CF] are concurrent.

PQRS is a parallelogram and T is a point inside the parallelogram such that the sum of \({\rm{P}}\hat {\rm{T}}{\rm{Q}}\) and \({\rm{R}}\hat {\rm{T}}{\rm{S}}\) is \({180^ \circ }\) . Show that \({\rm{TP}} \times {\rm{TR}} + {\rm{ST}} \times {\rm{TQ}} = {\rm{PQ}} \times {\rm{QR}}\) .

Since the lengths of the two tangents from a point to a circle are equal     (M1)

\({\rm{AF = AE, BF = BD, CD = CE}}\)     A1

Consider

\(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = 1\) (signed lengths are not relevant here)     M2A2

It follows by the converse to Ceva’s Theorem that [AD], [BE], [CF] are concurrent.     R2

[8 marks]

Draw the \(\Delta {\rm{PQU}}\) congruent to \(\Delta {\rm{SRT}}\) (or translate \(\Delta {\rm{SRT}}\) to form \(\Delta {\rm{PQU}}\) ).     R2

Since \({\rm{P}}\hat {\rm{U}}{\rm{Q}} = {\rm{S}}\hat {\rm{T}}{\rm{R}}\) , and \({\rm{S}}\hat {\rm{T}}{\rm{R + P}}\hat {\rm{T}}{\rm{Q}} = {180^ \circ }\) it follows that     R2  

\({\rm{P}}\hat {\rm{U}}{\rm{Q + P}}\hat {\rm{T}}{\rm{Q}} = {180^ \circ }\)     R1

The quadrilateral PUQT is therefore cyclic     R2

Using Ptolemy’s Theorem,     M2

\({\rm{UQ}} \times {\rm{PT}} + {\rm{PU}} \times {\rm{QT}} = {\rm{PQ}} \times {\rm{UT}}\)     A2

Since \({\rm{UQ = TR}}\) , \({\rm{PU = ST}}\) and \({\rm{UT = QR}}\)     R2

Then \({\rm{TP}} \times {\rm{TR}} + {\rm{ST}} \times {\rm{TQ}} = {\rm{PQ}} \times {\rm{QR}}\)     AG

[13 marks]