Let's do this question asks parallel to Eddie.
All right, and it's given that a B is equal to B.S. So these two sides are equal and that's equal.
Do it all right.
And if is equal to F.P..
So these two sides are equal.
If it's the midpoint of each find the ratio of any of EDC.
That's this triangle over here to the area of FGB.
That's this triangle over here.
All right.
Pause the video, give it a try and then let's do it together.
All right.
We're back.
Let's do it together.
I hope you have given it a try.
Now let's join point B to E and Andy.
All right, now let's start marking all the information that's already been given to us.
All right, let me do it again over here.
We know that Eddie and ACOR parallel, ABC and Eddie are equal and if is equal to FEMA.
But that's the information that we have.
All right, now.
We have joined Ebbe and Beatty right now.
In this way, we have divided this ship into six equal areas.
Right.
We had discussed this before.
In this way, we have divided this ship into six equal areas.
Now we need to find the ratio of any of EDC.
That's this part.
So for EDC, you have to pick one two.
You're taking two parts, right?
That's this part.
And this part two from EDC.
Now to find the area of F, a, B, you're just taking one part right there for the answer to this question
is to divide it by one.
Right.
This is two and this is one that's about it.
That is the answer.
Wasn't that fast.
Now, over here you can see that ECB is a parallelogram, right?
Let's try to understand it now.
Now, in the exam, you should have moved on to the next question.
Just marking.
It has to buy one.
All right.
Now, DCB is a battleground.
All right.
So area of triangle E, D, C, that's this part is equal to area of EBC.
Right.
All right, so these two are equal now you can see that these two are congruent, right?
We have proved that before.
Therefore, these two areas are also equal.
Now, if the area of DCB is taken as a then these two will be of area Ebeye to.
All right, now you can see that area of E. The beat, that's this part over here.
That is also equal to EBITA because this area and DBC is also equal right to area of EDB is also equal
to EBITA.
Now you can see that Eddy, be it this part over here also gives you Eppalock.
Right.
But this part also is a program.
So area of E, B, that is this part over here is equal to area of DB.
That is this part over here.
Let me write that over here.
These two areas are equal, and if we take the area of EDV as a then each of these is equal to by two.
All right.
Now, why have we taken the area of EDB?
Is it as well as the area of Iraq?
B, as A right.
What is the area of a program?
It's based in the height.
Right.
You will see that these two have the same base that's over here.
It's Abian over here.
It's busy.
And we know that they are of equal length.
And these are between two parallel lines.
So the height of these two also will be the same.
That's why both of them are of equal area.
All right.
Now what?
Over here you can see that the area of F a B is equal to half of EBITA.
Right, because these two triangles are congruent.
Therefore, the area of F a B is equal to Ebeye for now.
Having established these things, you can find that the required ratio is Ebeye two.
That's LDC, right.
EDC is equal to EBI two and F maybe that's it by four so ebi to buy by four and that gives you the answer
as to by one or two is to one.
And that's the same thing that we got within two or five seconds in the beginning itself, just using
graphical division method.