Question

) The chords AB and CD of the circle intersect at point M in the interier of
the same circle then prove that CM X BD = BM X AC.​

Answer 1

Given :- chords AB and CD of the circle intersect at point M in the interier of the same circle then prove that CM X BD = BM X AC.

Solution :- (from image .)

In ∆AMC and ∆DMB , we have,

→ ∠ACM = ∠DBM { Angles at the circumference subtended by the same arc are equal.}

similarly,

→ ∠CAM = ∠BDM .

so,

→ ∆AMC ~ ∆DMB { By AA similarity. }

then,

→ MC/MB = AC/DB

→ MC * DB = AC * MB

or,

→ CM * BD = BM * AC { Proved. }

Learn more :-

PQ and XY are two chords of a circle such that PQ = 6 cm and XY = 12 cm and PQ||XY. If the distance between the chords i...

https://brainly.in/question/24793483

PQRS is a cyclic quadrilateral with PQ = 11 RS = 19. M and N are points on PQ and RS respectively such that PM = 6, SN =...

https://brainly.in/question/27593946