Question

in the adjoining figure,CE is drawn parallel to DB to meet AB at E. Prove that:ar(quad.ABCD)=ar(triangle DAE)

Answer 1

Hey mate!!!

Here is your answer ;

In the given figure,

CE // DB,

So, we can tell that
ar(tri. DBC) =ar(tri. DBE)
[because they are on the same base DB and between the same parallels CE and DB.]

Therefore,

Adding ar(tri. ADB) both sides in the following equation, we get;

ar(tri. DBC) + ar(tri. ADB) = ar(tri. DBE) + ar(tri. ADB)

AE is a straight line.
Then,

=> ar(quad. ABCD) = ar(tri. ADE)


Here tri. stands for triangle and quad. stands for quadrilateral.


$<b ><big ><marquee >I hope it helps you #need more brainliests marks$

Answer 2

Answer:

hope it helps u well

Step-by-step explanation:

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