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In a parallelogram OABC , it is given OA = a , OC = c , PB = 1/3AB and BQ = 1/3 BC . OQ and AB are extended to meet at S while CB abd OP are extended to meet at R
Express (a) BS in terms of c
(b) BR in terms of a

(c) Show that RS is parallel to AC

In a parallelogram OABC , it is given OA = a , OC = c , PB = 1/3AB and BQ = 1/3 BC . OQ and AB are extended to meet at S while CB abd OP are extended to meet at R
Express (a) BS in terms of c
(b) BR in terms of a

(c) Show that RS is parallel to AC
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

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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

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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.