# P and Q are respectively the mid points of side AB and BC of a triangle ABC and R is the mid point of AP show that,{1} ar(PQR)= 1/2 ar(ARC){2} ar (RQC)=3/8 ar(ABC)3) ar (pBQ) = ar(ARC)

Given that: P and Q are the mid points of side AB and BC of a triangle ABC and R is the mid-point of AP

To prove:

Construction:  join PQ , RQ and RC

Proof:

(3)

CR is a median of

Also, CP is a median of

From (1) and (2), we get,

PQ is a median of,

From (3) and (4), we get,

Hence,

(1)

Since QP and QR medians of triangles QAB and QAP

And

From (6) and (7), we get,

From (5) and (8), we get,

Hence,

(2) Since CR is a median of,

…… (1)

Since RQ is median of,

Hence,

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let the area of triangle ABC be x.

(i) area (ΔPRQ)=1/2 area(APQ)[since RQ is a median and median divides triangle into equal areas]

= area(ΔABC)=............(1)

area (ΔARC)=1/2 area(ΔAPC) [in the ΔACP , CR is the median]

=1/2*1/2 area(ΔABC) [since CP is the median of ABC]

=1/4*x=x/4  ............(2)

from (1) and (2) area(ΔPRQ)=1/2 area(ΔARC)

(ii)

area(ΔRQC)=1/2 area(ΔARC) [since RQ is the median of triangle ARC]

=1/2*1/2area(ΔAPC)=1/4*1/2 area(ΔABC)=1/8x

(iii)

area(ΔPBQ)=area(ΔPQC)[the triangle with same base and between same set of parallel lines ]

=1/2 area(ΔAPC) [since PQ is the median of triangle APC]

=1/2*1/2 area(ΔABC)

=1/4 area(ΔABC)=x/4

from (2) area (ΔARC)=x/4

hence area(ΔPBQ)=area(ΔARC)

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