Question

Theorem of external division of chords.
If secants containing chords AB and CD of a circle intersect outside the circle in point E, then `AE xx EB = CE xx ED `
Given `:` (1) A circle with centre O
(2) Secants AB and CD intersect at point E outside the circle.
To prove `:` `AE xx EB = CE xx DE `
Construction `:` Draw seg AD and seg BC

Answer 1

Proof `:` `/_DAB ~= /_BCD` ….(Angles inscribed in the same arc are congruent )
i.e., `/_DAE ~= /_BCE ` …(A-B-E and C-D-E ) ….(1)
In `Delta ADE ` and `Delta CEB `,
`/_DAE ~= /_BCE` …[From (1)]
`/_AED ~= /_CEB ` ....(Common angle)
`:. Delta ADE ~ Delta CBE ` .....(AA test of similarity )
`:. (AE)/(CE ) = (DE)/( BE )` .... Corresponding sides of similar triangles are in proportion)
`:. AE xx BE = CE xx DE `.