Question

Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.

Answer 1

Let ABC be a triangle with A being the angle at the vertex and BC as the base. Let the mid points of AB be D and that of AC be E. Join DE and extend to F, such that EF= DE.

Consider $△ADE$ and $△CEF$

Statement Reason

AE = CE (E being the mid point of AC given ),

$\angle AED$ = $\angle CEF$ (vertically opposite angles)

DE = EF ( by construction.)

Hence $△ADE$ $\cong$ $△CEF$ [ By SAS congruency condition]

Thus AD = CF by cpctc.

Also AD = BD ( D is the midpoint of AB given) , Hence AD= CF= BD ( By transitive law)

Now consider quadrilateral DBCF

CF = BD ( From above)

CF II BD ( From above)

Hence Quad DBCF is a parallelogram ( The same pair of sides they are equal and parallel )

$\to$ DF = BC ( opp sides of a parallelogram)

$\to$ DE +EF = BC

$\to$ DE + DE =BC ( EF= DE from congruency condition above)

$\to$ 2DE = BC

$\to$ DE = $\frac{1}{2}$ BC

Proved