Question

prove that the line segment joining the midpoint of any two side of a triangle is parallel to the third side and is half of it. (by mid point theorem)

Answer 1

$\underline\mathfrak\green{Given:}$

In ∆ABC

D and E are two mid points of AB and AC

respectively

•°• AD = DB

AE = EC

$\underline\mathfrak\green{To\:Prove:}$

(i) DE ll BC

(ii) DE = $\bf\frac{1}{2}BC$

$\underline\mathfrak\green{Construction:}$

Draw lines 'l' through C such that L ll BA and produce DE to intersect 'l' at F.

$\underline\mathfrak\green{Proof:}$

(i) AE = EC (given)

$\angle$1 = $\angle$2 (vertically opp. angle)

$\angle$3 = $\angle$4 (alternate interior angles)

$\therefore$ ∆ADE $\congo$ ∆CFE (AAS)

1. DE = EF (cpct)
2. CF = AD (cpct)
3. Also, AD = BB (given)

From (2) and (3)

CF = DB

°•° CF = DB and CF ll DB (construction).

$\therefore$$\bf\underline{BCFD\:is\:a\:llgm}$

$\therefore$$\bf\underline{DF\:ll\:BC}$

$\bold{\large{\boxed{\sf{\pink{or,DE\:ll\:BC}}}}}$

(ii) DF = BC (BCFD is a llgm opp. sides of llgm are equal)

DE + EF = BC

2DE = BC

$\bold{\large{\boxed{\sf{\purple{or,DE=\frac{1}{2}BC}}}}}$