a) Express BS in terms of c.

Given: OC = c, and OA = a

So, BC = OA = a

BQ = (1/3)BC = a/3

CQ = BC -BQ = 2a/3

angle BQS = angle CQO .........vertical angles.

angle BSQ = angle COQ ........alternate interior angles.

Therefore, triangle BQS is similar to triangle CQO .....two angles of each triangle are equal each to each.

Similar triangles are proportional, by proportion,

OC/CQ = BS/BQ

c /(2a/3) = BS /(a/3)

Cross multiply,

c(a/3) = (BS)(2a/3)

c = 2BS

BS = c/2 --------------answer.

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a) Express BR in terms of a.

Given: OC = c, and OA = a

So, AB = OC = c

PB = (1/3)AB = c/3

AP = AB -PB = 2c/3

angle RPB = angle OPA .........vertical angles.

angle BRP = angle AOP ........alternate interior angles.

Therefore, triangle BRP is similar to triangle AOP.

By proportion,

OA/AP = BR/PB

a /(2c/3) = BR /(c/3)

Cross multiply,

a(c/3) = (BR)(2c/3)

a = 2BR

BR = a/2 --------------answer.

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(c) Show that RS is parallel to AC

angle RBS = angle CBA ...........vertical angles

BR/CB = BS/AB

(a/2)/a = (c/2)/c

1/2 = 1/2

True

Hence, triangle RBS is similar to triangle CBA.

And so, angle ACB = angle SRB

Therefore, RS is parallell to AC .....with RC as the transversal line, a pair of their alternate interior angles are congruent.