AB is a diameter and AC is a chord of the circle with centre at Q. the radius parallel to AC intersect the circle at D. Prove that D is the mid-point of the are BC.

Explanation

AB is diameter of the circle with centre at O and AC is a chord. The radius OD parallel to AC intersect the are BC at the point D.
We have to prove that Dis the mid-point of the are BC.
Construction : Let us join O,C
Proof : `:. AC||OD and OC` is the transversal of them.
`:. angle OCA= angle COD[ :.` alternative angle] .....(1)
Again `OA= OC[ :.` radii of same circle]
`:. angle OCA= angle OAC ......(2)`
Now, `angle BOC= angle BOD+ angle COD` [ by the figure]
or, internally opposit `(angle OCA+angle OAC)=angle BOD+angle COD`

or `angle COD+angle OAC= angle BOD+ angleCOD` [ by (1)]
`or, angle OAC= angle BOD or, angle COD= angle BOD` [ by (1)]
`:. angle BOD= angle COD` i.e, the two central angles produced by BD and CD are equal.
`:."are" BD="are" CD`.
Hence D is the mid-point of the are BC