MathLabs

Problem 4

Let PP be a point interior to triangle ABCABC (with CA≠CBCA \neq CB). The lines APAP, BPBP and CPCP meet again its circumcircle Γ\Gamma at KK, LL, respectively MM. The tangent line at CC to Γ\Gamma meets the line ABAB at SS. Show that from SC=SPSC = SP follows MK=MLMK = ML.
Step 3 of 5: Power of S produces a similarity
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SC2=SA⋅SB=SP2   ⟹   SPSA=SBSP   ⟹   △SPA∼△SBP   ⟹   ∠SAP=∠SPBSC^2=SA\cdot SB=SP^2\ \implies\ \dfrac{SP}{SA}=\dfrac{SB}{SP}\ \implies\ \triangle SPA\sim\triangle SBP\ \implies\ \angle SAP=\angle SPB
Detailed analysis

Since SCSC is tangent to Γ\Gamma at CC and S,A,BS,A,B are colinear, the power of SS with respect to Γ\Gamma equals both SC2SC^2 and SA⋅SBSA\cdot SB, so SC2=SA⋅SBSC^2=SA\cdot SB. Combined with the hypothesis SP=SCSP=SC, this gives SP2=SA⋅SBSP^2=SA\cdot SB, i.e. SP/SA=SB/SPSP/SA=SB/SP. Triangles SPASPA and SBPSBP share the angle at SS (angle ∠ASP=∠PSB\angle ASP=\angle PSB, the same angle at vertex SS between lines SA=SBSA=SB and SPSP), and the sides about it are proportional in this ratio, so by SAS similarity △SPA∼△SBP\triangle SPA\sim\triangle SBP. Matching corresponding angles of the similar triangles gives ∠SAP=∠SPB\angle SAP=\angle SPB.