$ABCD$ is a quadrilateral in which $P$,$Q$,$R$ and $S$ are mid-points of the sides $AB$,$BC$,$CD$ and $DA$ (see figure). $AC$ is a diagonal. Show that: $SR \parallel AC$ and $SR = \frac{1}{2} AC$.
(N/A) Given: $ABCD$ is a quadrilateral,$P$,$Q$,$R$,$S$ are mid-points of sides $AB$,$BC$,$CD$,$DA$ respectively. $AC$ is a diagonal.
To prove: $SR \parallel AC$ and $SR = \frac{1}{2} AC$.
Proof: In $\Delta ACD$,we have:
$S$ is the mid-point of $AD$.
$R$ is the mid-point of $CD$.
According to the Mid-point Theorem,the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it.
Therefore,in $\Delta ACD$,the line segment $SR$ joins the mid-points of sides $AD$ and $CD$.
Hence,$SR \parallel AC$ and $SR = \frac{1}{2} AC$.
To prove: $SR \parallel AC$ and $SR = \frac{1}{2} AC$.
Proof: In $\Delta ACD$,we have:
$S$ is the mid-point of $AD$.
$R$ is the mid-point of $CD$.
According to the Mid-point Theorem,the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it.
Therefore,in $\Delta ACD$,the line segment $SR$ joins the mid-points of sides $AD$ and $CD$.
Hence,$SR \parallel AC$ and $SR = \frac{1}{2} AC$.
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