Question

Pls solve the question ​

Answer 1

Answer:

SEG AC || SEG PD

SEG AC INTERSECTS THE TANGENT DRAWN AT C IN D

TO PROVE LINE BD IS A TANGENT TO THE CIRCLE

PROOF

IN ∆ APC

SEG CP = SEG AP. (radii of same circle)

L PAC = L PCA. 1(by isosceles

triangle theorem)

SEG AC || SEG DP AND CP IS THE TRANSVERSAL

L PCA= L CPD. 2(alternate angles)

L PAC = L CPD. 3(from 1,2)

AC || PD AND AP IS THE TRANSVERSAL

L PAC= L BPD. 4(corresponding angles)

L CPD = LBPD. 5(from 3,4)

IN ∆PCD ,∆PBD

SEG CP=SEG PB. (radii of same circle)

L CPD = LBPD. (from 5)

SEG PD= SEG PD

∆ PCD ~ ∆PBD. ( SAS)

L PCD = L PBD. ( c a c t)

L PBD = 90°

DB IS A TANGENT.

HENCE PROVED

HOPE IT'ILL HELP YOU......