Prove that PE * PF = PM * PK.

Solution:

• The problem asks to prove $$PE \cdot PF = PM \cdot PK$$.
• This equality suggests we should look for similar triangles involving these segments.
• Consider triangles PEM and PKF.
• Angle P is common to both triangles.
• We need another pair of equal angles.
• Angle PEM and angle PKF subtend arc MF. Thus, $$\angle PEM = \angle PKF$$.
• Therefore, by AA similarity, $$\triangle PEM \sim \triangle PKF$$.
• This similarity gives the proportion: $$\frac{PE}{PK} = \frac{PM}{PF}$$.
• Cross-multiplying this proportion gives: $$PE \cdot PF = PM \cdot PK$$.
• Alternatively, consider triangles PMF and PEK.
• Angle P is common to both.
• Angle PMF subtends arc KF. Angle PEK subtends arc KF. Thus, $$\angle PMF = \angle PEK$$.
• Therefore, by AA similarity, $$\triangle PMF \sim \triangle PEK$$.
• This similarity gives the proportion: $$\frac{PM}{PE} = \frac{PF}{PK}$$.
• Cross-multiplying this proportion gives: $$PM \cdot PK = PE \cdot PF$$.

Answer: Triangles PEM and PKF are similar by AA similarity because $$\angle P$$ is common and $$\angle PEM = \angle PKF$$ (angles subtended by the same arc KF). This similarity leads to the proportion $$\frac{PE}{PK} = \frac{PM}{PF}$$, which rearranges to $$PE \cdot PF = PM \cdot PK$$.