Question

in a triangle ABC, p is any point on bc .a line cq parallel to ap has been drawn which meet ba produced in q. prove that ar ( triangle bqp ) is equal to ar ( triangle ABC) ​

Answer 1

Answer:

To Prove :-RA = \frac{1}{3} AC

Construction :-

Draw PS parallel to BR to meet AC at S.

Proof :-

In Δ BCR, P is the mid-point of BC and PS is parallel to BR.

Where, S is the mid-point of CR

So, CS = SR ----- (1)

Again, In Δ APS, Q is the mid-point of AP and QR is parallel to PS.

Where, R is the mid-point of AS.

So, AR = RS ----- (2)

From equations (1) and (2),

We get, AR = RS = SC

⇒ AC = AR + RS + SC

⇒ AC = AR + AR + AR

⇒ AC = 3AR

∴ AR= \frac{1}{3} AC