Q15 of 56 Page 6

P and Q are the points on the sides AB and AC respectively of a ΔABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cms, QC = 6 cm, prove that BC = 3PQ.
                              

Given: Δ ABC, PQ are points on AB and AC such that AP = 2 cm, BP = 4 cm,  AQ = 3 cm, QC = 6 cm
To prove:  BC = 3PQ
 
Proof: In Δ ABC,
                     
As   AP = AQ
     PB     QC
According to converse of BPT, PQ || BC
   
In Δ APQ and Δ ABC 
∴ ∠ APQ = ∠ ABC (Corresponding angles)
∠ A is Common
∴ Δ APQ ~ Δ ABC (AAS similarity)
(corresponding sides of similar Δ s are proportional)
But
∴ PQ = 2 = 1
  BC    6    3
∴ 3PQ = BC (Proved).

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