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Triangles between Same Parallels

P and Q are r...

Question

P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP,
Show that $a r (Δ P B Q) = a r (Δ A R C)$ .

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Solution

Construction: Join PQ, RQ and RC

Proof:
(3)
CR is a median of
$Δ C A P$
$\Rightarrow a r (Δ A R C) = \frac{1}{2} a r (Δ C A P)$ ---- (1)
Also, CP is a median of
$Δ A B C$
$\Rightarrow a r (Δ C A P) = a r (Δ C B P)$ ------ (2)
From (1) and (2), we get,
$\Rightarrow a r (Δ A R C) = \frac{1}{2} a r (Δ C B P)$ ----- (3)
PQ is a median of
$Δ P B C$ ,
$\Rightarrow a r (Δ C P B) = 2 a r (Δ P B Q)$ ----- (4)
From (3) and (4), we get,
$\Rightarrow a r (Δ A R C) = a r (Δ P B Q)$ ----- (5)
Hence,
$a r (Δ A R C) = a r (Δ P B Q)$

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