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:
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).
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)