Application of the mid-point theorem - class-IX

In any triangle, the circumcentre, centroid, nine-point centre, and orthocentre all lie on a single straight line known as the Euler line, making the statement true.

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 $

Let the vertices be A, B, C. Midpoints are D(6,1), E(3,5), F(-1,-2). The vertex opposite to D is A. Using the property that the quadrilateral formed by midpoints is a parallelogram, A = E + F - D = (3-1-6, 5-2-1) = (-4, 2).

In an isosceles triangle with AB=AC, the median AD to the base BC is also the altitude. EF, connecting midpoints of AB and AC, is parallel to BC. Thus, AD is perpendicular to EF and bisected by it.

In an equilateral triangle with side 12, the altitude AD = 12 * sqrt(3)/2 = 6 * sqrt(3) approx 10.39. E is the midpoint of AD, so AE = 3 * sqrt(3). In right triangle ABE, BE^2 = AE^2 + AB^2 is not correct; rather, use triangle BDE where BD=6 and DE=3sqrt(3). BE^2 = 6^2 + (3sqrt(3))^2 = 36 + 27 = 63. BE = sqrt(63) approx 7.937.

The Midpoint Theorem states that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

The sum of the squares of the medians of a triangle is 3/4 times the sum of the squares of the sides. (3/4) * 32 = 24.

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.

For a right triangle with sides 9, 12, 15, the sum of squares of sides is 81 + 144 + 225 = 450. The sum of squares of medians is (3/4) * 450 = 337.5.

In $\triangle ABC$

The angle bisectors meet at the incenter, the medians meet at the centroid, and the perpendicular bisectors of the sides meet at the circumcenter. All three statements are standard geometric properties.

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

The triangle formed by joining the midpoints of the sides of a triangle has an area equal to 1/4 of the area of the original triangle. (1/4) * 4 = 1.