Q : 7     P and  Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that

             (i)  \small ar(PRQ)=\frac{1}{2}ar(ARC)

Answers (1)


We have \DeltaABC such that P, Q and R are the midpoints of the side AB, BC and AP respectively. Join PQ, QR, AQ, PC and RC as shown in the figure.
Now, in \Delta APC,
Since R is the midpoint. So, RC is the median of the \DeltaAPC
Therefore,  ar(\DeltaARC) = 1/2 . ar (\DeltaAPC)............(i)

Also, in \DeltaABC, P is the midpoint. Thus CP is the median.
Therefore, ar(\DeltaAPC) = 1/2. ar (\DeltaABC)............(ii)

Also, AQ is the median of \DeltaABC 
Therefore, 1/2. ar (\DeltaABC) = ar (ABQ)............(iii)

In \DeltaAPQ, RQ is the median.
Therefore, ar (\DeltaPRQ) = 1/2 .ar (\DeltaAPQ).............(iv)

In \DeltaABQ, PQ is the median
Therefore, ar(\DeltaAPQ) = 1/2. ar(\DeltaABQ).........(v)


From eq (i),
\frac{1}{2}ar(\Delta ARC)= \frac{1}{4}ar(\Delta APC)...........(vi)
Now, put the value of ar(\DeltaAPC) from eq (ii), we get 
Taking RHS;
= \frac{1}{8}[ar(\Delta ABC)] 
= \frac{1}{4}[ar(\Delta ABQ)]        (from equation (iii))

= \frac{1}{2}[ar(\Delta APQ)]   (from equation (v))

 =ar(\Delta PRQ) (from equation (iv))

\Rightarrow\small ar(PRQ)=\frac{1}{2}ar(ARC)

Hence proved.

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