Question

the adjoining figure, AB is the
diameter of circle with centre O, AC
is tangent at point A and BC
intersects the circle at point D. Line
JH touches the circle at point D and
intersects AC in point J.
Prove that seg AJ = seg CJ.​

Answer 1

Answer:

Given AB is the  diameter , AC  is tangent at point A and BC  intersects the circle at point D

In ΔDOC and ΔBOD

⇒OC=OB    ....(Radii of same circle)

⇒CD=DB   ....(Tangent from an external point)

⇒OD=OD    ....(Common side)

⇒ΔCOD=ΔBOD      (SSS test)

⇒ΔCOD=ΔBOD=x     (c.a.s.t.) equation (i)

In ΔAOC,  AO=OC   ....(Radii of same circle)

⇒  ∠OAC=∠OCA=y  ....(Isoceles Δ theorem) equation (ii)

⇒∠COB=∠OAC+∠OCA   ....(Remote interior ∠ theorem)

⇒∠COD+∠BOD=∠OAC+∠OCA

⇒x+x=y+y   ....[From equation (i)  and (ii)]

⇒2x=2y

⇒x=y   ....(dividing by 2)

⇒∠COD=∠OCA

⇒  Seg OD∥ chord AC.   ....(Alternate ∠s test)              

Similarly Seg AJ=Seg CJ   ....(Alternate ∠s test)  

Hence, proved.

Answer 2

Answer:

Step-by-step explanation: