Question

if secants containing chords AB and CD of a circle intersect outside the circle in point E, then AE×EB = CE×ED

Answer 1

Firstly we have THAT AE =CE--1
AS SECANTS FROM SAME EXTERNAL POINT.

AND SIMILARLY EB=ED---2

MULTIPLY 1 AND 2

WE WILL GET
$ae \times eb = ce \times ed$

Answer 2

Answer:

In ∆ADE and ∆CBE, ∠AED = ∠CEB [Common angle] ∠DAE ≅ ∠BCE [Angles inscribed in the same arc] ∴ ∆ADE ~ ∆CBE [AA testof similaritv] ∴ AE/CE = ED/EB [Corresponding sides of similar triangles] ∴ AE × EB = CE × EDRead more on Sarthaks.com - https://www.sarthaks.com/854240/if-secants-containing-chords-and-circle-intersect-outside-the-circle-in-point-then-ae-eb-ce