Question

Prove that a line drawn through the mid point of one side of a triangle parallel to another side bisects the third side ( Using basic proportionality theorem )​

Answer 1

$\underline\red{\underline{\sf{\maltese\: Given:-}}}$

Let us assume ∆ABC.

Where DE is parallel to BC & D is the mid point of AB.

$\underline\red{\underline{\sf{\maltese\: To\: prove:-}}}$

E is the mid point of AC

$\underline\red{\underline{\sf{\maltese\: Proof:-}}}$

In ∆ABC, DE II BC

If a line drawn parallel to one side of triangle, intersects the other two sides in distinct points, then it divides the other 2 side in same ratio.

$\red\implies$$\tt{\frac{AD}{DB} = \frac{AE}{EC}}$

$\red\implies$$\tt{\frac{DB}{DB} =\frac{AE}{EC}}$

[As, D is the midpoint of AB. So, AD = DB]

$\red\implies$ $\tt{ 1= \frac{AE}{EC}}$

EC = AE

$\red\implies$ E is the mid-point of AC.

(Since D is the mid-point of AB)

Hence proved :)

Answer 2

Answer:

$\underline\red{\underline{\sf{\maltese\: Given:-}}}$

✠Given:−

Let us assume ∆ABC.

Where DE is parallel to BC & D is the mid point of AB.

$\underline\red{\underline{\sf{\maltese\: To\: prove:-}}}$

✠Toprove:−

E is the mid point of AC

$\underline\red{\underline{\sf{\maltese\: Proof:-}}}$

✠Proof:−

In ∆ABC, DE II BC

If a line drawn parallel to one side of triangle, intersects the other two sides in distinct points, then it divides the other 2 side in same ratio.

$\red\implies⟹ \tt{\frac{AD}{DB} = \frac{AE}{EC}}$

$DB</p><p>AD</p><p> = </p><p>EC</p><p>AE$

$\red\implies⟹ \tt{\frac{DB}{DB} =\frac{AE}{EC}}$

$DB</p><p>DB </p><p> = </p><p>EC</p><p>AE$

[As, D is the midpoint of AB. So, AD = DB]

$\red\implies⟹ \tt{ 1= \frac{AE}{EC}}1= </p><p>EC</p><p>AE$

$EC = AE$

$\red\implies⟹ E is the mid-point of AC.$

(Since D is the mid-point of AB)

Hence proved :)