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Chord set

If two chords intersect in a circle, the product of the two intercepts on one chord is equal to the product of the intercepts on the other chord - my homework . If two chords intersect in a circle, the product of the two segments on one chord is equal to the product of the segments on the other chord.

Proof of the chord theorem

Prerequisite:

AC and BD are two chords in the circle. S is the intersection of the chords.

Assertion:

AS(average)⋅SC(average)=BS(average)⋅SD(average)

Proof:

The distances AB and CD are also chords of the circle, so ∢ BDA=∢ BCA and ∢ DBC=∢ DAC (circumferential angle theorem) holds and triangle ASD is similar to triangle CSB (principal similarity theorem).

Thus holds:

AS(average):BS(average)=SD(average):SC(average) so also AS(average)⋅SC(average)=BS(average)⋅SD(average) (w. z. b. w.)

For the special case where one chord is diameter of the circle - geometry problem solver - and the other chord is perpendicular to it, there is a right triangle because, according to Thales' theorem, any circumferential angle over the diameter is a right angle.

Angle on a circle

An angle is called a center angle if its vertex lies at the center of the circle, a circumference angle if its vertex lies on the circle and its legs intersect the circle, and a chord-tangent angle if its vertex lies on the circle and one leg intersects the circle and the other touches the circle.

An angle is called a center angle - homework help geometry (centric angle) if its vertex is at the center of the circle, circumferential angle (peripheral angle) if its vertex is on the circle and its legs intersect the circle, chord-tangent angle if its vertex is on the circle and one leg intersects the circle and the other touches the circle .

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