Mid-point theorem states that

Mid-Point Theorem : 
The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the third side.

On applying the midpoint theorem, we get
 $DE = \cfrac {AC}{2} = \cfrac 72 = 3.5$cm

$ ABCD $ is a rhombus. $ AB =BC = CD = DA $

using formula incenter $x= \dfrac{ax _{1}+bx _{2}+cx _{3}}{a+b+c}$ 

Given: $\triangle ABC$, P is mid point of BC, $QR \parallel BC$ and $PQ \parallel AC$

Since, $ PQ \parallel AC$ and P is mid point of BC, thus, by converse of mid point theorem
Q is mid point of AB.

Now, In $\triangle ABP$
Since, $QR \parallel BP$ and Q is mid point of AB. thus, by converse of Mid point theorem
R is mid point of AP.
Hence, $QR = \frac{1}{2} BP$ (Mid point theorem)
$QR = \frac{1}{2} (\frac{1}{2} BC)$ (P is midpoint of BC)
$BC = 4 QR$

Given: D and F are mid points of AB and AC respectively.
Hence, by mid point theorem, $DF \parallel BC$

Also, given $BD \parallel EF$
Since, opposite sides are parallel to each other. Hence, $BDEF$ is a parallelogram

Perimeter of BDEF = $2 (BD + BF)$ (Opposite sides of parallelogram are equal)
Perimeter of BDEF = $AB + BC$ (D and F are mid points of AB and BC respectively)
Perimeter of BDEF = $16 + 18$
Perimeter of BDEF = $34$ cm

In $\triangle ABC$, $M$ is mid point of $AB$ and $N$ is mid point of $AC$
$D$ is any point of BC
Now, Join AD and MN such that they met at O
In $\triangle ABC$
M is mid point of AB and N is mid point point of AC
Hence, $MN \parallel BC$ and $MN = \frac{1}{2} BC$

Now, In $\triangle ABD$
$MO \parallel BC$ and M is mid point of AB
Thus, $O$ is mid point of AD
Hence, $MN$ bisects $AD$

Given: $ABCD$ is a rhombus. $P, Q, R, S$ are mid points of $AB, BC, CD, DA$ respectively.
Join: $AC$ and $BD$

In $\triangle ABC$
$P$ is mid point of $AB$ and $Q$ is mid point of $AC$.
Thus, by mid point theorem, $PQ \parallel AC$ and $PQ = \dfrac{1}{2} AC$
Similarly, In $\triangle ACD$,
$S$ is mid point of $AD$ and $R$ is mid point of $CD$.
Thus, by mid point theorem, $SR \parallel AC$ and $SR = \dfrac{1}{2} AC$
Hence, $PQ \parallel SR$ and $PQ = SR$

Similarly, $PS = QR$ and $PS \parallel QR$
Thus, the opposite sides of $PQRS$ are equal and parallel.
We know the diagonals of a rhombus bisect each other at right angles.
Now, since $AC \perp BD$ thus, $PS \perp PQ$ (Angle between two lines is same as the angle between their corresponding parallel lines)

Now, the opposite sides of $PQRS$ are equal and parallel and the sides meet each other at right angles. Hence, $PQRS$ is a rectangle.

If $PQRS$ had to be a square, the diagonals must be equal and bisect at right angles. It is possible only if $ABCD$ is a square.

Let the point on the line $x+y=25$ be $P(a,b)$
Thus equation of chord of contact $AB$ from point $P$ to the circle is given by,
$T  =0 \Rightarrow ax+by = 9$  (i)
Let mid point of $AB$ be $R(h,k)$.
Now equation of chord $AB$ with mid point $R$ is given by,
$T = S _1 \Rightarrow hx+ky = h^2+k^2$ (ii)
Both line (i) and (ii) represents the same line $AB$
$\therefore \displaystyle \frac{a}{h}=\frac{b}{k} = \frac{9}{h^2+k^2}$
$\Rightarrow  a=\cfrac{9h}{h^2+k^2}, b = \cfrac{9k}{h^2+k^2}$
Also point $(a,b)$ lie on the line $x+y = 25$
$\Rightarrow a+b = 25\Rightarrow \cfrac{9h}{h^2+k^2}+ \cfrac{9k}{h^2+k^2}=25$
$ \Rightarrow 25(h^2+k^2) = 9(h+k)$
Hence required locus of $R(h,k)$ is given by $25(x^2+y^2) = 9(x+y)$

In $\triangle ABC$

Given in $\triangle ABC$, $D,E,F$ are the mid points of sides $AB,BC$ and $CA$ respectively

No. of ways $={^{ 3 }{ { C } _{ 1 } }}\times {^{ 4 }{ { C } _{ 1 }} }\times {^{ 5 }{ { C } _{ 1 } }}+{^{ 3 }{ { C } _{ 2 } }}\left( ^{ 4 }{ { C } _{ 1 } }+{^{ 5 }{ { C } _{ 1 } }}\right)+{^{ 4 }{ { C } _{ 2 }} } \left(^{ 3 }{ { C } _{ 1 } }+{^{ 5 }{ { C } _{ 1 }} }\right) +{^{ 5 }{ { C } _{ 2 }} }\left( ^{ 3 }{ { C } _{ 1 } }+{^{ 4 }{ { C } _{ 1 } }}\right)$