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प्रश्न
AB and CD are two chords of a circle intersecting at P. Prove that AP x PB = CP x PD
उत्तर
Construction: Join AD and CB.
In ΔAPD and ΔCPB
∠A = ∠C .....(Angles in the same segment)
∠D = ∠B .....(Angles in the same segment)
⇒ ΔAPD ~ ΔCPB ...(By AA Postulate)
`=> (AP)/(CP) = (PD)/(PB)` ...(Corresponding sides of similar triangles)
`=> AP xx PB = CP xx PD`
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संबंधित प्रश्न
In the figure given below, O is the centre of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD.
AB = 24 cm, OM= 5 cm, ON= 12 cm. Find the:
1) radius of the circle
2) length of chord CD.
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In the given figure, 5 × PA = 3 × AB = 30 cm and PC = 4 cm. Find CD.
In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.
Prove that, of any two chords of a circle, the greater chord is nearer to the centre.
Chords AB and CD of a circle intersect each other at point P such that AP = CP
Show that: AB = CD
In two concentric circles, prove that all chords of the outer circle, which touch the inner circle, are of equal length.
Two tangents are drawn to a circle from an external point P. touching the circle at the points A and B. A third tangent intersects segment PA in C and segment PB in D and touches the circle at Q. if PA = 20 units, find the perimeter of Δ PCD.
If AB and CD are two chords which when produced meet at P and if AP = CP, show that AB = CD.
Two circles of radii 10 cm and 8 cm intersect and the length of the common chord is 12 cm. Find the distance between their centres.
