Question

In the figure, $D, E$ and $F$ are midpoints of sides $AB, BC$ and $AC$ respectively. $P$ is the foot of the perpendicular from $A$ to side $BC$. Show that points $D, E, F$ and $P$ are concyclic.

Answer 1

Hint:
Here, we will use the midpoint formula to prove the quadrilaterals formed inside the triangle is a parallelogram. Then we will use the properties of parallel lines and hence, find the relationship between various angles. Solving further and proving the quadrilateral $PDFE$ as a cyclic quadrilateral will help us to show that the points $D,E,F$ and $P$ are concyclic.

Complete step by step solution:
Given: We are given a figure such that in $\vartriangle ABC$, $D,E$ and $F$ are midpoints of sides $AB,BC$ and $AC$ respectively. And, $P$ is the foot of the perpendicular from $A$ to side $BC$.
To prove: $D,E,F$ and $P$ are concyclic.
Proof: In $\vartriangle ABC$, $D$ and $F$ are the mid-points of the sides $AB$ and $AC$ respectively.
Therefore, by mid-point theorem,
$DF\parallel BC$
Similarly, by mid-point theorem,
$EF\parallel AB$
And, $ED\parallel CA$
Now, in quadrilateral $BEFD$
$DF\parallel BE$($\because DF\parallel BC$)
And, $EF\parallel BD$($\because EF\parallel AB$)
Hence, since the opposite sides of this quadrilateral are parallel.
Thus, this is a parallelogram.

Similarly,
Quadrilateral $ADEF$ is also a parallelogram.
Hence, we can say that,
$\angle A = \angle DEF$( This is because opposite angles of a parallelogram are equal)
Now, we can see that \[ED\parallel AC\] and $EC$ is the transversal
Hence,
$\angle BED = \angle C$ (Corresponding angles)
Hence, we can say,
\[\angle BEF = \angle DEB + \angle DEF{\text{ }} = \angle C{\text{ }} + \angle A\]……………………..$\left( 1 \right)$
Similarly,
$DF\parallel BC$ and $BD$ is the transversal,
Hence,
$\angle ADF = \angle B$ (Corresponding angles)………………………………..$\left( 2 \right)$
Now,
In $\vartriangle ABP$, $D$ is the mid-point of $AB$ and $OD\parallel BP$
Hence, by converse of mid-point theorem,
$O$ is the mid-point of $AP$
Hence, we get,
$OA = OP$
Now, in $\vartriangle AOD$ and $\vartriangle DOP$
$OA = OP$ (Proved above)
$\angle AOD = \angle DOP = 90^\circ $ (Linear pair)
And, $OD = OD$ (Common)
Hence, $\vartriangle AOD \cong \vartriangle DOP$ (By Side Angle Side, SAS congruency)
Hence, $\angle ADO = \angle PDO$ (By CPCT)
From $\left( 2 \right)$, we know that,
$\angle ADF = \angle B$
Here, $\angle ADO = \angle ADF = \angle PDO$
Therefore,
$\angle PDO = \angle B$……………………………..$\left( 3 \right)$
Now, in quadrilateral $PDFE$
$\angle PDO + \angle PEF = \angle B + \left( {\angle A + \angle C} \right)$ ( From $\left( 1 \right)$and $\left( 3 \right)$)
But, by angle sum property of triangles, we know that,
$\angle A + \angle B + \angle C = 180^\circ $
Therefore,
$\angle PDO + \angle PEF = \angle A + \angle B + \angle C = 180^\circ $
Hence, clearly, the sum of opposite angles of the quadrilateral $PDFE$ is $180^\circ $
Hence, quadrilateral $PDFE$ is a cyclic quadrilateral.
Therefore, the points $D,E,F$ and $P$ are concyclic.
Hence, verified
Hence, this is the required answer.

Note:
According to the mid-point theorem, it states that the line segment in a triangle joining the mid-points of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side. Whereas, the converse of the mid-point theorem states that if a line is drawn through the midpoint of one side of a triangle and is parallel to the other side, then, it bisects the third side. Hence, the mid-point theorem played an important role in solving this question.