In the figure given below, AF = DF and AB // FE // DC.

Prove that :

(i) $\text{FP} = \dfrac{1}{2}\text{ AB}$

(ii) AB + CD = 2 EF

(i) Given: AF = DF and AB // FE // DC.

To prove: $\text{FP} = \dfrac{1}{2}\text{ AB}$

Proof: In Δ FDP and Δ ADB,

∠ADB = ∠FDP (Common angle)

∠DAB = ∠DFP (corresponding angles since AB // FE)

∠DBA = ∠DPF (corresponding angles since AB // FE)

By AAA similarity rule,

Δ FDP ~ Δ ADB

So, $\dfrac{FP}{AB} = \dfrac{FD}{AD}$

As it is given that AF = FD,

FD = $\dfrac{1}{2}$ AD

So, $\dfrac{FP}{AB} = \dfrac{1}{2}$

Hence, $\text{FP} = \dfrac{1}{2}\text{ AB}$.

(ii) To prove: AB + CD = 2 EF

Proof: From (i), $\text{FP} = \dfrac{1}{2}\text{ AB}$

⇒ 2FP = AB ………………….(1)

In Δ BEP and Δ DCB,

∠DBC = ∠PAE (Common angle)

∠BPE = ∠BDC (corresponding angles since AB // FE)

∠BEP = ∠BCD (corresponding angles since AB // FE)

By AAA similarity rule,

Δ BEP ~ Δ DCB

So, $\dfrac{EP}{DC} = \dfrac{BE}{BC}$

As it is given that AF = DF and AB // FE // DC, so BE = EC

BE = $\dfrac{1}{2}$ BC

So, $\dfrac{EP}{DC} = \dfrac{1}{2}$

⇒ 2EP = DC ………………….(2)

Adding equation (1) and (2), we get

⇒ 2FP + 2EP = AB + DC

⇒ 2(FP + EP) = AB + DC

⇒ 2FE = AB + DC

Hence, AB + CD = 2 EF .