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# 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) ar (PRQ) = 1/2 ar (ARC)

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

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We have $\Delta$ABC 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 $\Delta$APC
Therefore,  ar($\Delta$ARC) = 1/2 . ar ($\Delta$APC)............(i)

Also, in $\Delta$ABC, P is the midpoint. Thus CP is the median.
Therefore, ar($\Delta$APC) = 1/2. ar ($\Delta$ABC)............(ii)

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

In $\Delta$APQ, RQ is the median.
Therefore, ar ($\Delta$PRQ) = 1/2 .ar ($\Delta$APQ).............(iv)

In $\Delta$ABQ, PQ is the median
Therefore, ar($\Delta$APQ) = 1/2. ar($\Delta$ABQ).........(v)

From eq (i),
$\frac{1}{2}ar(\Delta ARC)= \frac{1}{4}ar(\Delta APC)$...........(vi)
Now, put the value of ar($\Delta$APC) 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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