Question

Prove that in a triangle the line segment joining the
midpoints of any two sides is parallel to the third side
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given :-

A triangle ABC such that D is the midpoint of AB and E is the midpoint of AC.

To Prove :-

DE || BC and DE = 1/2 BC

Construction :-

Through C, draw a line CF || AB, intersecting DE when produced to F.

Proof :-

As it is given that,

D is the midpoint of AB

$\rm\implies \:AD = DB$

E is the midpoint of AC

$\rm\implies \:AE = EC$

Now,

$\rm \: In \: \triangle \:AED \: and \: \triangle \:CEF$

$\rm \: AE = EC \: \{given \}$

$\rm \: \angle \:1 \: = \: \angle \:2 \: \: \{vertically \: opposite \: angles \}$

$\rm \: \angle \:3 \: = \: \angle \:4 \: \: \{alternate \: interior \: angles \}$

$\rm\implies \:\triangle \:AED \: \cong \: \triangle \:CEF \: \: \{ASA \: congruency \: rule \}$

$\rm\implies \:DE \: = \: EF \: \: \: \: \: \{CPCT \}$

and

$\rm\implies \:AD \: = \: CF \: \: \: \: \: \{CPCT \}$

As, it is given that,

$\rm \: \:AD = DB$

So,

$\rm \: \:CF = DB - - - (1)$

Also, By Construction, we have CF || AB

$\rm \: \:CF \: \parallel \: DB - - (2)$

From equation (1) and (2), we concluded that

$\rm \: \:CF = DB \: \: and \: \: CF \: \parallel \: DB$

We know, In a quadrilateral, if opposite pair of sides are equal and parallel, then quadrilateral is a parallelogram.

$\rm\implies \:CFDB \: is \: a \: parallelogram$

$\rm\implies \:\rm \: BC \: = \: FD \: \: and \: BC \: \parallel \: FD$

$\rm\implies \:\rm \: DE + EF \: = \: BC \: \: and \: \: DE \parallel \:BC$

$\rm\implies \:\rm \: DE + DE \: = \: BC \: \: and \: \: DE \parallel \:BC$

$\bigg[\rm\because \:DE \: = \: EF \: \: \: \: \: \{proved \: above \} \: \bigg]$

$\rm\implies \:\rm \: 2DE \: = \: BC \: \: and \: \: DE \parallel \:BC$

$\rm\implies \:\rm \: DE \: = \: \dfrac{1}{2} \: BC \: \: and \: \: DE \parallel \:BC$

Hence, Proved

$\rule{190pt}{2pt}$

Answer 2

solution :

• please check the attached file