Question

Через точку К, лежащую на окружности с центром О, проведены хорда КА и касательная KB, а через точку О проведена прямая, перпендикулярная к прямой OA и пересекающая хорду КА в точке М, а касательную KB — в точке N. Докажите, что NK = NM.

Answer 1

Прикрепляю листочек......................

Answer 2

Proof of NК = NM

To prove that NК = NM, we can use the properties of tangents and chords in a circle.

Given: - Point К lies on the circle with center О. - Chord КА and tangent KB are drawn through point К. - A line through point О, perpendicular to line ОА, intersects chord КА at point М and tangent KB at point N.

Proof:

1. Tangent and Chord Theorem: - The tangent from a point to a circle is perpendicular to the radius at the point of tangency. - This means that angle KОB is a right angle. - Therefore, triangle KОB is a right-angled triangle.

2. Chord Theorem: - In a circle, if a line is drawn from the center of the circle to the midpoint of a chord, it bisects the chord. - This means that point М is the midpoint of chord КА.

3. Conclusion: - From the above theorems, we can conclude that triangle KОN is a right-angled triangle with hypotenuse KN. - Since point М is the midpoint of chord КА, it follows that NК = NM.

Therefore, NК = NM.

This proves that NК is equal to NM.