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..
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
In the fig. ABCD is a parallelogram and AC is the diagnol.
To prove - - AB=CD nd AD= BC
consider triangle ABD nd BCD
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
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
Consider triangle ABD nd triangle BCD
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,
angle A + angle B = angle C+ angle D --- 3
angle A + angle B + angle C+ angle D = 360o (angle sum property of a quadrilateral)
FRom 3 v hve, angle A + angle B = angle C+ angle D
angle A + angle B + angle A + angle B = 360o
2(angle A + angle B) = 360o
angle A + angle B = 360o/ 2
angle A + angle B = 180o -- 4
angle C = angle D =180o ---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..
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.