Solve the previous question using the Baudhāyana–Pythagoras theorem.
Given: CE ⊥ AB, CH ⊥ FG, and CE = CH. Since the perpendicular from the centre bisects the chord, E and H are the midpoints of AB and FG, respectively. Applying the Baudhāyana–Pythagoras theorem to right triangles ΔCEA and ΔCHF, we get: CA² = CE² + AE² and CF² = CH² + FH².
Since CA and CF are radii of the same circle, CA = CF. Therefore, CE² + AE² = CH² + FH². We are given that CE = CH, which means CE² = CH². Substituting this into our equation, we get AE² = FH², which implies AE = FH.
Because E and H are midpoints, AB = 2AE and FG = 2FH. Since AE = FH, we can conclude that AB = FG. Hence, chords equidistant from the centre have equal length.
Explanation
The question asks to prove Theorem 7 (chords equidistant from the centre have equal length) using the Baudhāyana–Pythagoras theorem. The context provides the logic for Theorem 8 which uses this theorem to compare distances of unequal chords. By applying the theorem to the right triangles formed by the radius, perpendicular distance, and half the chord, we can equate the squares of the radii. Since the distances from the centre are equal, their squares cancel out, leaving the squares of the half-chords equal, thus proving the chords are equal.
Solution Steps
Step 1: Apply the Baudhāyana–Pythagoras theorem to right ΔCEA and ΔCHF: CA² = CE² + AE² and CF² = CH² + FH².
Step 2: Equate the expressions for CA² and CF² since CA = CF (radii), resulting in CE² + AE² = CH² + FH².
Step 3: Use the given condition CE = CH to cancel CE² and CH², leaving AE² = FH², which means AE = FH.
Step 4: Since E and H are midpoints, AB = 2AE and FG = 2FH. Because AE = FH, it follows that AB = FG.