Question

In the adjoining figure seg ab is a diameter of a circle with centre o. The bisector of angle abc intersects the circle at point d prove that seg ad is congruent seg bd

Answer 1

AD is congruent BD , seg AD  is congruent seg BD

Step-by-step explanation:

∠AOD  = 2 ∠ACD  ( angle formed by same chord AD at center & arc segment)

∠ACB = 90°   ( as AB is Diameter)

CD is bisector of ∠ACB

=> ∠ACD = ∠BCD = 90°/2

=> ∠AOD  = 2 (90°/2 )

=> ∠AOD  =90°

=> ∠BOD  = 180° - 90° = 90°

now in ΔAOD  & ΔBOD

OA = OB  ( radius)

OD = OD  Common

∠AOD  =  ∠BOD  = 90°

=> ΔAOD  ≅ ΔBOD

=> AD = BD

AD is congruent BD

∠AOD  =  ∠BOD  = 90°

=> seg AD  is congruent seg BD

QED Proved

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