Mid-point Theorem and Medians - Class IX

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.

In a right triangle ABC, using the Pythagorean theorem for triangles ABE and BCD, AE^2 = AB^2 + BE^2 and CD^2 = BC^2 + BD^2. Substituting BE = BC/2 and BD = AB/2, we get AE^2 + CD^2 = 5/4(AB^2 + BC^2), which equals 5/4(AC^2).

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. 

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

In triangle ABD, AD^2 = AB^2 + BD^2. Since D is the midpoint of BC, BD = BC/2. Thus, AD^2 = AB^2 + (BC/2)^2, which simplifies to AD^2 = AB^2 + BC^2/4. Rearranging gives BC^2 = 4(AD^2 - AB^2).

This is Menelaus' Theorem, which states that for a line intersecting the sides of a triangle, the product of the ratios of the segments is 1.

Using Apollonius theorem on triangle ABC with median AD, 2(AD^2 + BD^2) = AB^2 + AC^2. With BE as a median to AC, 2(BE^2 + AE^2) = AB^2 + BC^2. Solving these equations with given values leads to AC = 2*sqrt(7).

The length of a median m_b is given by m_b = 1/2 * sqrt(2a^2 + 2c^2 - b^2). Rearranging this formula for side b gives b = 2/3 * sqrt(2m_a^2 + 2m_c^2 - m_b^2).

Given AB = A1B1 and M, N, M1, N1 are midpoints, the vectors MM1 and NN1 represent the displacement between the midpoints of the sides of two congruent triangles, which must be equal.

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.

By geometric construction and properties of the point P, the conditions given satisfy the requirements for A, P, and N to be collinear.

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.

A line joining the midpoints of opposite sides of a parallelogram is parallel to the other two sides and divides the parallelogram into two smaller parallelograms of equal area.

The vectors AD, BE, and CF are medians of the triangle. The sum of the vectors from the vertices to the midpoints of the opposite sides is zero.