Circles
Power of a Point Theorem
Grade 11
Question:
<p>Let ABCD be a rectangle with diagonals AC and BD intersecting at O. A straight line through B intersects DC produced at E and DA produced at F such that OE = OF. Prove or disprove: \((CE)(DE) = (AF)(DF)\).</p>
<p>(A) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1</p>
<p>(B) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1</p>
<p>(C) Statement-1 is true, Statement-2 is false</p>
<p>(D) Both statements are false</p>
Step-by-Step Solution
Key Concept: Since OE = OF and O is the center of the rectangle, points E and F lie on a circle centered at O. The power of point B with respect to this circle gives the required equality.
<p><strong>Analysis:</strong> Since O is the center of rectangle ABCD, it is equidistant from all vertices. The condition OE = OF means E and F lie on a circle centered at O. By the power of point B with respect to this circle: \((BE)(BQ) = (PB)(BR)\) where the line through B intersects the circle at E and F. This gives \((CE)(DE) = (AF)(DF)\). Statement-2 correctly applies the power of a point theorem.</p>
Correct Answer: A