For any chord through A, its length depends on its distance from O. The chord is shortest when this distance is greatest. This occurs when the chord is perpendicular to OA. Hence proved.
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In the figure, radii divide the opposite angles into parts. Using the isosceles triangles formed by equal radii, we get p + v = 90° and q + u = 90°. Hence, opposite angles sum to 180°.
As written, this statement appears inconsistent with the figure/text. Since each chord is perpendicular to diameter AB, its midpoint lies on AB. Therefore, the segment joining the two midpoints lies along AB, so it is parallel to AB, not perpendicular.
In the figure, the diameter passes through O and A lies on the semicircle. Since OA, OB and OC are radii, the relevant triangles are isosceles. Their angles show that a + b = 90°, so the angle at A ...
The diameter passes through the centre and is the longest chord of a circle. Any other chord lies away from the centre and is therefore shorter than the diameter. Hence, no chord is longer.