Question

$\vartriangle PQR$ is an equilateral triangle in which $M$ is the midpoint of $QR$ and $N$ is the midpoint of $MR$.
Prove that $P{N^2} = 13N{R^2}$

Answer 1

Hint: We are given an equilateral triangle in which a line is perpendicular to one side of triangle and it divide that side into two equal parts and another line which divides other equal part into two equal parts as we know that sides of an equilateral triangle are equal and the median divides the sides into two equal parts as the altitude is given then in a right angle triangle we use Pythagoras theorem
It states that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse (the side opposite the right angle) or, in familiar algebraic notation
$c^2$=$a^2$+$b^2$
Here $a = $base, b=height and c=hypotenuse
And then form the equation using it and then arrange it accordingly to the equation given to prove.

Complete step-by-step answer:
Step1: We are given an equilateral triangle in which $M$ is the midpoint of $QR$ and $N$ is the midpoint of $MR$.
As $\vartriangle PQR$ is equilateral
$\therefore PQ = QR = PR$
$M$ is the midpoint of$QR$
Hence $QM = MR$
Step2: Now we will apply Pythagoras theorem in triangle $\vartriangle PQM$
C2 = a2 +b2
Here c$ = PQ$, a$ = PM$, b$ = QM$
$\Rightarrow$$P{Q^2} = P{M^2} + Q{M^2}$…(1)
As we know $QM = MR$ substituting this in (1)
$\Rightarrow$$P{Q^2} = P{M^2} + M{R^2}$
On further solving we get
$\Rightarrow$$P{Q^2} - M{R^2} = P{M^2}$…(2)
In $\vartriangle PMN$ we apply Pythagoras theorem
$P{M^2} = P{N^2} - M{N^2}$
As $PN$ is the median
$MN = NR$(N is the midpoint of $\vartriangle PMR$)
Substituting the value of $MN$ as $NR$
$\Rightarrow$$P{M^2} = P{N^2} - N{R^2}$
Step3: Substituting the value of $P{M^2}$ from equation (2)
$\Rightarrow$$P{Q^2} - M{R^2} = P{N^2} - N{R^2}$
As triangle is equilateral hence $PQ = QR$
Substituting $QR$ in place of $PQ$
$Q{R^2} - M{R^2} = P{N^2} - N{R^2}$
$\therefore MR = 2NR$($N$ is the midpoint of $MR$)
Substituting the value of $MR$
$\Rightarrow$$Q{R^2} - {(2NR)^2} = P{N^2} - N{R^2}$…(3)
Step3: As $2MR = QR$…(4) (as $M$ is the midpoint)
And $N$ is the midpoint of $MR$ hence
$MR = 2NR$
Putting this value of $MR$ in equation (4)
$\Rightarrow$$2(2NR) = QR$
$\Rightarrow$$4NR = QR$
Substituting this value of $QR$ in equation (3)
$\Rightarrow$${(4NR)^2} - {(2NR)^2} = P{N^2} - N{R^2}$
$\Rightarrow$$16N{R^2} - 4N{R^2} = P{N^2} - N{R^2}$
On arranging like terms on one side we get:
$\Rightarrow$$16N{R^2} - 4N{R^2} + N{R^2} = P{N^2}$
$\Rightarrow$$13N{R^2} = P{N^2}$
Hence it is proved that $13N{R^2} = P{N^2}$

Note: In such types of questions students in the beginning are not able to make the approach to it. In such cases first see the equation to prove then mainly in right angled Triangles use the Pythagoras Theorem form the equation using it and then substitute the value in those equations according to the required equation to prove.