Q) Diagonals AC and BD of quadrilateral ABCD intersects at O . If AB=CD , then show that 1) ar(DOC) = ar(AOB) 2) ar (DCB) = ar(ACB) [ question from NCERT class 9 textbook.

In triangle ABD , AO is the median therefore ar(AOB) = ar(AOD) [ since median of a triangle divides it into 2 triangles with equal areas ]...................................(1) and also in triangle BCD , CO is the median and ar(COB) = ar ( COD) ...................................(2) . Adding (1) and (2) we get ar(ABD)= ar(BCD) . Thus diagonal divides it into 2 congruent triangles. Now we consider equal lines AB and CD after joining their end points we would either get an isosceles trapezium if the lines are not parallel or a parallelogram if the lines are parallel . But the diagonal of a isos. trapezium does not divide it into 2 triangles with equal areas , therefore it must be a parallelogram . Now triangles ACB and BCD are on the same base BC and between same parallels AD and BC . therefore ar(DCB)=ar(ACB) [ proof 2] and subtracting ar(BOC) from both sides we get ar(DOC)=ar(AOB] [ proof 1 ] . Hence proved . [ The question can easily be done by drawing perpendiculars from B and D on AC.]

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest[Adding (1) and (2) we get ar(ABD)= ar(BCD) ].................................... this part of your answer goes wrong because this part can be only be prooved if bo would be the median if BO would be the median than TRIANGLE BOC AND BOA would have equal area. and then ar[cob]=ar[boa]. Then area of all the four triangles would be equal. then you can prove your first part. so, this prove is wrong.

Log in to reply

But , I have not used BO . I have just used median AO and CO .

Log in to reply

but by using ao and co u can not prove ar[abd] =ar [dcb]. i think you are not clear with my answer let me explain . let us take ar[abd]=x and ar[cbd]=y then since ao is the median of x ar [aod]=ar[aob]=1/2x. since co is the median ar[cod]=ar[cob]=1/2y. and 1/2y would be only equal to 1/2x if their areas are equal that is do would be the median.i hope your doubt would be clear by this for more doubt about this contact me at 0788(2290223).

Log in to reply

@karandeep singh ludhar @abhishek anand .. Kaddu's way is correct.. I did it the same way. This question appeared in our exams too right?.. Kaddu apna e-mail id de

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply