ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).

pls explain the fiure.....sure for thumbs up for good answer..

#### Answers

In the fig. ABCD is a trapezium with AB II DC.. nd

COnstruction- Draw AC (digonal) . Draw a line XY parallel to AC in which X intersects at AB nd Y intersects at BC

__ COnsider triangle ADX nd triangle ACX __

since these triangles share a common base AX nd lie between same parallels AB nd DC..

ar(ADX) = ar(ACX) ---- 1

__ Consider triangle ACX nd triangle ACY __

since dese triangles share common base AX and lie between same parallels AC nd XY

ar(ACX)= ar(ACY) --- 2

From 1 and 2

__ ar(ADX) = ar(ACY) __

hope u undrstud diss.. if any problem plss tell me!!!

hey akshita, from where do u cut nd past all the answers???

hii i did dis ans on my own.. nd its not copied..sorry :)))

and can u explain how tro draw this figure accordinf to given information..m confused paaa plsss sure for thumbs up

m sorry shwetha for lte reply

first draw trapezium abcd.. thn draw the diagonal AC nd then draw XY parallel to AC.. u cn find it wen u do dis X intersects at AB nd Y at BC..

if u didnt still undrstnd plss ask mee..

then waht abt DC AND CX.its joined...y..pls answer..

2) in a parallelogram , opposite sides are equal..prove this pls..sure for thumbs up 2morrow exam..

3) if each ppair of opposite sides of a quadrilateral is equal , then its is a prallelogram,

4) if in a quadrilateral , each pair of opp . angles is equal, then it is a parallelogram..

5) if the diagonals of a quadrilateria;s bisect each other , then it is II gm ..

id ont know to do the converse of the theorem...can u suggeast me a good idea..plzzzzzzzzz

ya i got the answer for 5) one could u pls ...do me 2 , 3 , 4 ,plss

hii shwetha.. DC is a part of trapezium.. nd CX u hve to join in order to prove the triangles r equal

2)

In the fig. ABCD is a parallelogram and AC is the diagnol.

To prove - - AB=CD nd AD= BC

Proof

consider triangle ABD nd BCD

- angle ADB= angle CBD ( alternate interior angle)
- AC = AC (common)
- angle ABD= angle CDB ( alternate interior angle)

therefore triangle ADC congruent to triangle ABC (by ASA congruence rule)

ther4 AB= CD (by cpct)

AD = BC(by cpct)

ther4 opp sides of parallelogram r equal

3)

In the fig. ABCD is a quadrilateral with diagonal AC.

Given -- AB=CD and AD= BC

To prove - ABCD is a parallelogram, since opp sides r equal, we only hve to prove AB II CD nd AD II BC

Proof

Consider triangle ABD nd triangle BCD

- AB = CD(given)
- AD=BC(given)
- BD=BD(common)

therefor triangle ABD congruent to triangle BCD (by SSS congruence rule)

ther4 angle ADB=CBD(by cpct).. wich implies AD II BC

angle ABD=CDB(by cpct).. wich implies AB II CD

**SInce the opp sides of a quadrilateral r equal nd parallel.. the gvn quadrilateral is a parallelogram..**

. ...

IN the fig. ABCD is a quadrilateral ...

Given - angle A = angle C ---1

and angle B = angle D --- 2

Adiing equations 1 and 2, 1+2,

we get

angle A + angle B = angle C+ angle D --- 3

We have

angle A + angle B + angle C+ angle D = 360^{o} (angle sum property of a quadrilateral)

FRom 3 v hve, angle A + angle B = angle C+ angle D

therfore

angle A + angle B + angle A + angle B = 360^{o}

2(angle A + angle B) = 360^{o}

angle A + angle B = 360^{o}/ 2

angle A + angle B = 180^{o} ^{ }-- 4

angle C = angle D =180^{o} ^{ }---5

From 4.. AD II BC (since <A nd <B r cointerior angles of the quadrilateral... if the sum of cointerior angles r supplementary thn the sides r parallel to each odr)

Similarly frm 5 AB II CD

**Hence ABCD is a parallelogram...**

this one is liitle bit confusinfg where did u get the pic...diagram..

frm net

hey akshita can u plzz say us d site frm wer u get dese answers plzzz !!!!!

sorry to say this akshita but when u wrote

__ ' 'Consider triangle ACX nd triangle ACY __

since dese triangles share common base AX ....... ' '

the base isnt AX it is AC.