Question

if two equal chords of a circle intersect within the circle prove that the segments of one chord are
equal to the corresponding segment of the code​

Answer 1

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Answer 2

Answer:

Proved

Step-by-step explanation:

if two equal chords of a circle intersect within the circle prove that the segments of one chord are

equal to the corresponding segment of the code​

Let say two chord AB & CD intersect at E

AB = CD = x

to prove AE = CE   & BE = DE

We will draw an perpendicular at chords from center of circle O

at chord AB at point M  &  at Chord CD at point N

Perpendicular from center at chord bisect the chord

AM = AB/2 = x/2  

CN = CD/2 = x/2

OM²  = Radius² - (x/2)² & ON²  = Radius² - (x/2)²

=> OM² = ON²

we will draw a line from center O to point E

we will get two right angle triangles

ΔOEM & ΔOEN

ME² = OE² - OM²

NE² = OE² - ON²

OM² = ON²

so ME² = NE²

=> ME = NE

AE = AM + ME

CE = CN + NE

AM = CN  & ME = NE

=> AE = CE

BE = AB - AE

DE = CD - CE

AB = CD & AE = CE

so

BE = DE

QED