Question

In ΔABC, the mid-points of AB, BC and AC are P, Q and R respectively. Prove that BQRP is a parallelogram and that its area is half of ΔABC.

Answer 1

Since P and R are mid-points of AB and AC respectively.

Therefore, PR || BC and PR = `(1)/(2)"BC"` ...........(i)

Also Q is mid-point of BC,

⇒ QC = `(1)/(2)"BC"`      ...........(ii)

From (i) and (ii)
PR || BC and PR = QC
⇒ PR || QC and PR = QC     ..........(iii)
Similarly Q and R are mid-point of BC and AC respectively
Therefore, QR || BP and QR = BP   ..........(iv)
⇒ PQ is a digonal of ||gm BQRP
ar(ΔPQR) = ar(ΔBQP)  ....(v) (diagonal of a ||gm divides it into two triangles of equal areas)
Similarly QCRP and QRAP are || gm and
ar(ΔPQR) = ar(ΔQCR) = ar(ΔAPR)    ..........(vi)
From (v) and (vi)
ar(ΔPQR) = ar(ΔBQP) = ar(ΔQCR) = ar(ΔAPR)
Now,
ar(ΔABC) = ar(ΔPQR) + ar(ΔBQP) + ar(ΔQCR) + ar(ΔAPR)
⇒ ar(ΔABC) = 4ar(ΔPQR)

⇒ ar(ΔPQR) = `(1)/(4)"ar(ΔABC)"`   ..........(vii)

ar(||gm BQRP) = ar(ΔPQR) + ar(ΔBQP)
⇒ ar(||gm BQRP) = ar(ΔPQR) + ar(ΔPQR)    ...(from(v))
⇒ ar(||gm BQRP) = 2ar(ΔPQR)
⇒ ar(||gm BQRP) = `2 xx (1)/(4)"ar(ΔABC)"` ...(from(vii))
⇒ ar(||gm BQRP) = `(1)/(2)"ar(ΔABC)"`.