The side AC of a triangle ABC is produced to point E so that CE = $\dfrac{1}{2}AC$. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meet AC at point P and EF at point R respectively. Prove that :

(i) 3DF = EF

(ii) 4CR = AB.

By mid-point theorem,

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.

By converse of mid-point theorem,

The straight line drawn through the mid-point of one side of a triangle parallel to another, bisects the third side.

By equal intercept theorem,

If a transversal makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts.

(i) In △ ABC,

D is the mid-point of BC and DP || AB.

∴ P is the mid-point of AC. (By converse of mid-point theorem)

∴ AP = PC

Given,

⇒ CE = $\dfrac{1}{2}AC$

∴ CE = PC

Since, AP = PC and CE = PC,

∴ AP = PC = CE.

Since,

⇒ AB || DP || CR

⇒ AF || DP || CR (Since, point F lies on straight line AB)

In △ AEF,

AF || PD || CR and AP = PC = CE

∴ DF = DR = RE = x (let) [By equal intercept theorem]

From figure,

⇒ EF = DF + DR + RE = x + x + x = 3x = 3DF.

Hence, proved that 3DF = EF.

(ii) In △ ABC,

D is mid-point of BC and DP || AB.

∴ P is the mid-point of AC. (By converse of mid-point theorem)

∴ PD = $\dfrac{1}{2}AB$ (By mid-point theorem) ………(1)

In △ PED,

Given,

⇒ CE = $\dfrac{1}{2}AC$

⇒ CE = PC (Since, P is mid-point of AC)

∴ C is the mid-point of PE.

C is mid-point of PE and DP || CR.

∴ R is the mid-point of DE. (By converse of mid-point theorem)

∴ CR = $\dfrac{1}{2}PD$ (By mid-point theorem) ………(2)

Substituting value of PD from equation (1) in equation (2), we get :

⇒ CR = $\dfrac{1}{2} \times \dfrac{1}{2}AB$

⇒ CR = $\dfrac{1}{4}AB$

⇒ 4CR = AB.

Hence, proved that 4CR = AB.