1. A line XY is drawn parallel to the side BC of the $$\begin{gathered} ∆ABC,and its cuts AB and AC at X and Y respectively.If BE\parallel AC and CF\parallel AB \\ are drawn to cut XY at E and F respectively ,prove that ar\left(∆ABC\right) =ar\left(∆ACF\right). \\ 2.D is the middle point of the side AB\hspace{0.17em}of the ∆ABC ,P is any point on the side BC.If CQ\parallel PD \\ is drawn to cut AB at Q then prove that ar\left(∆BPQ\right)=\frac{1}{2}ar\left(∆ABC\right) \end{gathered}$$

They have given that EB||AC , AB||FC and also XY||BC.
To Prove : ar(ABE)=ar(ACF).
Construction : Extend YX to Intersect EB at M and extend XY to intersect CF at N. So that, EA||MY and AF||XN.
Proof : 
            EB||AC and AB is transversal, 
    ->>angle "CAB=EBA"
​            Also, EA||BC and AB is transversal, (MY||EA & MY||BC .So,EA||BC)
    ->>angle "ABC=EAB"

In triangle ABC and EAB,
    ->>angle"CAB=EBA"
    ​->>angle"ABC=EAB"
    ->>  AB=AB (common side)
By ASA criteria,
    ->>triangle ABC congruent to triangle EAB.->1
Similarly,
    ​->>triangle ABC congruent to triangle ACF.->2

From 1 & 2,
     ->>triangle ABE congruent to triangle ACF...
 
So,      ar(ABE)=ar(ACF)

     ->> Hence proved!!!
 HOPE IT HELPS YOU