Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true.
Given: Circle with centre C, chords AB = FG. E and H are midpoints of AB and FG respectively. CE and CH are perpendiculars from C to AB and FG.
To show: CE = CH.
By the Baudhāyana–Pythagoras theorem in right triangles CEA and CHF, we have CA² = CE² + AE² and CF² = CH² + FH². Since CA and CF are radii of the same circle, CA = CF, which means CA² = CF². Therefore, CE² + AE² = CH² + FH². Since E and H are midpoints of equal chords AB and FG, AE = FH, which means AE² = FH². Substituting this, we get CE² = CH², which implies CE = CH. Thus, chords of equal length are equidistant from the centre.
Explanation
The question asks to prove Theorem 6 using the Baudhāyana–Pythagoras theorem instead of congruence. Theorem 6 states that equal chords are equidistant from the centre. By dropping perpendiculars from the centre to the chords, we form two right triangles. Applying the Pythagoras theorem to both, and equating the squares of the radii (since they are equal), we can substitute the equal half-chord lengths to prove that the squares of the perpendicular distances are equal, hence the distances themselves are equal.
Solution Steps
Step 1: Identify right triangles formed by dropping perpendiculars from centre C to chords AB and FG, forming CE ⊥ AB and CH ⊥ FG.
Step 2: Apply Baudhāyana–Pythagoras theorem: CA² = CE² + AE² and CF² = CH² + FH².
Step 3: Equate CA² and CF² since CA and CF are radii: CE² + AE² = CH² + FH².
Step 4: Substitute AE² = FH² (since E and H are midpoints of equal chords AB and FG).
Step 5: Conclude CE² = CH², which gives CE = CH.