AB = AC
Hence, $\angle ABC = \angle ACB$ (Isosceles triangle property)
Now, since, D and F are mid point of AB and AC respectively, thus DF II BC (Mid point theorem)
Hence, 
$\angle ADF = \angle ABC$ and $\angle AFD = \angle ACB$ (Corresponding angles)
Thus, $\angle ADF = \angle ABC = \angle AFD = \angle ACB$ 
Now, In $\triangle$ ADF and FEC
$\angle ADF = \angle FEC$ (Corresponding angles of parallel lines EF and AB)
$\angle AFD = \angle ACB $(Corresponding angles of parallel lines DF and BC)
AF = FC (F is the mid point of AC)
Thus $\triangle ADF \cong \triangle FEC$ (AAS rule)
Hence, AD = FE (corresponding sides of congruent triangles)
Similarly, we can prove, AF = DE
Since, AD = AF (half lengths of equal sides, AB and AC)
Thus, EF = DE or $\triangle$ DEF is an isosceles triangle.

The mid point is $(\dfrac{a-5a}2,\dfrac{b+7b}2)$.i.e $(-2a,4b)$
Option D is correct.

Medians : The line segment from any vertex of a triangle to the midpoint of its opposite side is called medians of triangle. 

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Line joining the mid-points of any two sides of a triangle is parallel to the third side. 

$\begin{array}{l}AD = \sqrt {2A{B^2} + 2A{C^2} - B{C^2}}  = 6\2A{B^2} + 2A{C^2} - B{C^2} = 36\BE = \sqrt {2A{B^2} + 2B{C^2} - A{C^2}}  = 8\2A{B^2} + 2B{C^2} - A{C^2} = 64\CF = \sqrt {2A{C^2} + 2B{C^2} - A{B^2}}  = 10\2A{C^2} + 2B{C^2} - A{B^2} = 100\A{B^2} = x\A{C^2} = y\B{C^2} = z\2x + 2y - z = 36\2x + 2z - y = 64\2y + 2z - x = 100\x = A{B^2} = \frac{{100}}{9}\y = A{C^2} = \frac{{208}}{9}\z = B{C^2} = \frac{{292}}{9}\AD = 6,BE = 8,CF = 10\in,\Delta ABE\AD \bot BE\area,\Delta BEC = 16\area,\Delta ABE = 16 + 16 = 32\\frac{{area,\Delta ABE}}{{area,\Delta DEF}} = 4\\frac{{32}}{{area,\Delta DEF}} = 4\area,\Delta DEF = 8\end{array}$

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 $x+y = 25$
$\Rightarrow a+b = 25 \Rightarrow 25(h^2+k^2) = 9(h+k)$
Hence required locus of $R(h,k)$ is given by, $25(x^2+y^2) = 9(x+y)$

Since M is a midpoint on AB and N is a midpoint on AC, we have

$ AB = AM

+ MB $   and $ AB = AN + NC $

Since, $

AB = AC $

$ =>

AM + MB = AN + NC $

$

=>  AM = AN $

M is the mid point of AB and MN II BC. Thus, N is the mid point of AC and

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, $AP = 2 AR$

Given: $D$ is mid point of $AB$ and $E$ is mid point of $BC$, $F$ is any point on $AC$ and $EF \parallel AB$

Now, in $\triangle ABC$,
E is mid point of BC and $EF \parallel AB$
By Mid point Theorem, $F$ is mid point of $AC$

Also, D is mid point of AB and F is mid point of AC
Hence, by mid point theorem, $DF \parallel BE$
Since, $DF \parallel BE$ and $EF \parallel AB or BD$
Hence, BEFD is parallelogram.

Given that $E$ and $F$ are mid points. We have to prove that ${a} _{1}(BOC)={a} _{1}(AEOF)$

Given: 

The triangle ABC has height h and base l,
E is the midpoint of BC, line parallel to the base will be interect the triangle at the midpoint of the opposite side.
Triangle ADE is similar to triangle ABC, as all 3 angles are equal.
Therefore, the altitude of ADE is half of ABC, and DE will be half of BC.

Given: In $\triangle ABC, D, E$ and $F$ are midpoints of sides $AB, BC$ and $CA$.

Let $(x,y)$ be the coordinates of B.
According to question, M is mid point of A and B.
$\Rightarrow \dfrac{-2+x}2=1\Rightarrow x=4$
and $ \dfrac{3+y}2=0\Rightarrow y=-3$
Therefore B is $(4,-3)$
Option B is correct.

Fact: Using mid-point theorem $\dfrac{\Delta{DEF}}{\Delta{ABC}}=4$

The middle piece is the tubular structure in which mitochondria are spirally arranged and it also has the beginning part of the flagellum. The sperm tail or the flagellum is based upon unique 9+2 arrangement. This arrangement refers to the nine peripheral, symmetrically arranged microtubule doublets.
Thus, the cross-section of the middle piece of sperm will show mitochondria and 9+2 arrangement of microtubules.
Hence, the correct answer is option (C), 'Mitochondria and 9+2 arrangement of microtubules'.