Question

in the figure , sides AB, BC, CA Of triangle ABC are produced upto points R, P, F respectively such that AB=BR, BC=CP and CA=AF Prove that area of triangle PFR is equal to 7 area of triangle ABC

Answer 1

Given: Sides AB, BC and CA of Δ ABC are produced up to points R, P and F respectivelysuch that AB = BR, BC = CP, and CA = AF

Construction: Join PA, FB and RC.



Let ar(Δ ABC) = a

We know that median of a triangle divides it into triangles of equal area.

In Δ PAB, AC is the median

⇒ ar(Δ ABC) = ar(Δ APC) = a

In Δ PCF, PA is the median

⇒ ar(Δ APC) = ar(Δ PFA) = a

In Δ BCF, BA is the median

⇒ ar(Δ ABC) = ar(Δ ABF) = a

In Δ RAF, FB is the median

⇒ ar(Δ ABF) = ar(Δ RBF) = a

In Δ ACR, BC is the median

⇒ ar(Δ ABC) = ar(Δ RBC) = a

In Δ PRB, RC is the median

⇒ ar(Δ RBC) = ar(Δ RPC) = a

Thus, ar(Δ ABC) = ar(Δ APC) = ar(Δ PFA) = ar(Δ ABF) = ar(Δ RBF) = ar(Δ RBC) = ar(Δ RPC) = a

Hence, ar(Δ PFR) = ar(Δ ABC) + ar(Δ APC) + ar(Δ PFA) + ar(Δ ABF) + ar(Δ RBF) + ar(Δ RBC) + ar(Δ RPC)

= a + a + a + a + a + a + a = 7a = 7 ar(Δ ABC)