Theorem of internal division of chords. Suppose two chords of a circle intersect each other in the interior of the circle, then the product of the lengths of the two segments of one chord is equal to the product of the lengths of the two segments of the other chord. Given `:` (1) A circle with centre O . (2) chords PR and QS intersect at point E inside the circle. To prove `:` PE `xx` ER = QE xx ES ` Construction `:` Draw seg PQ and seg RS

Theorem of internal division of chords. Suppose two chords of a circle

1.	Theorem of internal division of chords. Suppose two chords of a circle intersect each other in the interior of the circle, then the product of the lengths of the two segments of one chord is equal to the product of the lengths of the two segments of the other chord. Given `:` (1) A circle with centre O . (2) chords PR and QS intersect at point E inside the circle. To prove `:` PE `xx` ER = QE xx ES ` Construction `:` Draw seg PQ and seg RS
Answer» Proof `:` `/_PQS ~= /_SRP ` ....(Angles inscribed in the same arc ar congruent ) i.e., `/_PQE ~= /_SRE ` ….(P-E-R and Q-E-S ) ….(1) In `Delta PQE` and `Delta SRE`, `/_PQE ~= /_SRE` ….[From (1) ] `/_ PEQ ~= /_SER` .....(Vertically opposite angles ) `:. Delta PQE ~ Delta SRE ` .....(AA test of similarity ) ` :. ( PE )/( SE) = (QE)/(RE)` ......(Corresponding sides of similar triangles are in proportion ) `:. PE xx ER = QE xx SE `.