In a triangle ABC, P and Q are respectively the mid points of AB and BC and R is the midpoint of AP. Prove that:

Question:

In a triangle ABC, P and Q are respectively the mid points of AB and BC and R is the midpoint of AP. Prove that:

(i) ar(ΔPBQ) = ar(ΔARC).

(ii) ar(ΔPRQ) = 1/2 ar(ΔARC).

(iii) ar(ΔRQC) = 3/8 ar(ΔABC).

Solution:

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

(i) Since CR is the median of ΔCAP

∴ ar(ΔCRA) = (1/2) ar(ΔCAP)   ⋅⋅⋅⋅⋅ (1)

Also, CP is the median of a Δ CAB

∴ ar(ΔCAP) = ar(ΔCPB)  ⋅⋅⋅⋅ (2)

From 1 and 2, we get

∴ ar(ΔARC) = (1/2) ar(ΔCPB)  ⋅⋅⋅⋅⋅  (3)

PQ is the median of a ΔPBC

∴ ar(ΔCPB) = 2ar(ΔPBQ)   ⋅⋅⋅ (4)

From 3 and 4, we get

∴ ar(ΔARC) = ar(ΔPBQ)   ⋅⋅⋅⋅ (5)

(ii) Since QP and QR medians of triangles QAB and QAP respectively

∴ ar(ΔQAP) = ar(ΔQBP)   ⋅⋅⋅⋅  (6)

And ar(ΔQAP) = 2ar(ΔQRP) ⋅⋅⋅⋅ (7)

From 6 and 7, we get

ar(ΔPRQ) = (1/2) ar(ΔPBQ) ⋅⋅⋅⋅⋅ (8)

From 5 and 8, we get

ar(ΔPRQ) = (1/2) ar(ΔARC)

(iii) Since, LR is a median of ΔCAP

∴ ar(ΔARC) = (1/2) ar(ΔCAD)

= 1/2 × (1/2) ar(ΔABC)

= (1/4) ar(ΔABC)

Since RQ is the median of ΔRBC.

∴ ar(ΔRQC) = (1/2) ar(ΔRBC)

= (1/2) {ar(ΔABC) − ar(ΔARC)}

= (1/2) {ar(ΔABC) – (1/4) ar(ΔABC)}

= (3/8) ar(ΔABC)

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