Given: pk and pf tangent to circle k(o) k and f points of tangency

prove: m∠p=m∠ek

See explanation

Step-by-step explanation:

If PK and PF are tangent to the circle, then the radii OK and OF are perpendicular to the lines PK and PF respectively.

Consider quadrilateral PFOK. The sum of the measures of all interior angles in quadrilateral PFOK is equal to 360°. Since OK and OF are perpendicular to the lines PK and PF, you have that

∠FOK+90°+∠KPF+90°=360°,

∠FOK=360°-90°-90°-∠KPF=180°-∠KPF.

Angles FOK and KOE are supplementary angles, then

∠KOE=180°-∠FOK=180°-(180°-∠KPF)=∠KPF.

Since angle KOE is central angle, the arc EK has the same measure as angle KOE and as angle KPF.

quit

step-by-step explanation:

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval