Question

Please answer bro please please

Answer 1

We know that

"Triangles lying on same or equal bases and between the same parallel, have equal areas"

So according to the theorem,

ar(∆ABC) = ar(∆DBC)

Now, we can see that,

ar(∆ABC) = ar(∆PAB) + ar(∆PBC)

and ar(∆DBC) = ar(∆PDC) + ar(∆PBC)

but,

ar(∆ABC) = ar(∆DBC)

=> ar(∆PAB) + ar(∆PBC) = ar(∆PDC) + ar(∆PBC)

cancel ∆PBC from both sides,

=> ar(∆PAB) = ar(∆PBC)

Hence Proved.

Answer 2

$\textbf{\huge{ANSWER:}}$

There's a theorem:

The triangles on the same base and between the same parallel lines are equal in areas.

According to the same theorem,
ar (Triangle ABC) = ar (Triangle DBC)

Triangle PBC is common in both the given triangles. Thus, if we subtract it's area, the rest area will also be equal. Doing the same:

=》 ar (Triangle ABC) - ar (Triangle PBC) = ar (Triangle DBC) - ar (Triangle PBC)

=》 ar (Triangle PAB) = ar (Triangle PDC)

Hence Proved!

Hope it Helps!! :)