In ΔPQO, DE || OQ (Given)

∴ PD/DO = PE/EQ ...**(i)** [By using Basic Proportionality Theorem]

In ΔPQO, DE || OQ (Given)

∴ PD/DO = PF/FR ...**(ii)** [By using Basic Proportionality Theorem]

From equation **(i)** and **(ii)**, we get

PE/EQ = PF/FR

In ΔPQR, EF || QR. [By converse of Basic Proportionality Theorem]

In the fig 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

In ΔOPQ, AB || PQ (Given)

∴ OA/AP = OB/BQ ...**(i)** [By using Basic Proportionality Theorem]

In ΔOPR, AC || PR (Given)

∴ OA/AP = OC/CR ...**(ii)** [By using Basic Proportionality Theorem]

From equation **(i)** and **(ii)**, we get

OB/BQ = OC/CR

In ΔOQR, BC || QR. [By converse of Basic Proportionality Theorem].

Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side.

Given: ΔABC in which D is the mid point of AB such that AD=DB.

A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.

To Prove: E is the mid point of AC.

Proof: D is the mid-point of AB.

∴ AD=DB

⇒ AD/BD = 1 ... **(i)**

In ΔABC, DE || BC,

Therefore, AD/DB = AE/EC [By using Basic Proportionality Theorem]

⇒1 = AE/EC [From equation **(i)**]

∴ AE =EC

Hence, E is the mid point of AC.

Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side

Given: ΔABC in which D and E are the mid points of AB and AC respectively such that AD=BD and AE=EC.

To Prove: DE || BC

Proof: D is the mid point of AB (Given)

∴ AD=DB

⇒ AD/BD = 1 ... **(i)**

Also, E is the mid-point of AC (Given)

∴ AE=EC

⇒AE/EC = 1 [From equation **(i)**]

From equation **(i)** and **(ii)**, we get

AD/BD = AE/EC

Hence, DE || BC [By converse of Basic Proportionality Theorem]