Geometry help?

Question

ABCD is a parallelogram. BC is extended to a point Q such that CQ = AD. AQ intersects CD at P. Join DQ. Prove that ar (BPC) = ar(DPQ)
{Hint: Join AC}

No trigonometry allowed. I need answers that use basic principles in NCERT textbook Grade 9.

Answer 1

look, the answer comes very simply. the diagram should be big n neatly made.

first prove DPQ=APC(bases are same.heights equal.or by congruency);
then prove,,,APC=BPC(triangles lying between two parallel lines have same area);
hence...DPQ=BPC

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i know in this much may b it will not be very clear,mail me if u want the diagram or any other help regarding this question

Answer 2

in fact, the sole factor that i'm able to discover in common with sacred geometry and alchemy is they the two manage styles of spirituality. Sacred geometry deals with using geometric types and shapes interior the layout and shape of church homes, temples, mosques or everywhere of religious gathering. those particular shapes mean particular issues, based on the religion or concept gadget. to illustrate, the pentagram in Christianity refers back to the 5 senses. In Catholicism, it may each and every each and every now and then seek advice from the devil, or the Evil One. Alchemy deals with the technique of turning metals into gold, applying chemical compounds and technological awareness. in spite of the incontrovertible fact that, Alchemy is likewise a concept in looking the appropriate know-how by way of chemistry. hence, alchemy is a spirituality all its own. All in all, the sole factor that i'm able to work out in common between the two is they the two manage a form, or types, of spirituality.

Answer 3

ok i think i hav got the answer.
given:parallellogram abcd.AD=BC
BC extended to Q such that AD=CQ
and AQ intersects CD at P
to prove:ar(bpc)=ar(dpq)
constructions:drop a perpendicular from Q to PC meeting PC at M.
drop a perpendicular from from B to meet PC extended at F
proof:ar(DPQ)=1/2 *base*altitude
=1/2 *DP*QM........(1)
ar(BPC)=1/2 * PC*BF.........(2)

now, QM=QC sin BF=BC sin QC=AD(GIVEN)
QC=BC (AD=BC)
Hence,QM=BF.......(3)



ACDQ is another parallellogram
CD AND AQ ARE ITS DIAGONALS
IN A PARALLELLOGRAM DIAGONALS BISECT EACH OTHER
HENCE DP=PC.........(4)
FROM (3) AND (4)
(1)=(2)
HENCE PROVED

Answer 4

When all four sides of your parallelogram ABCD is equal, then this can be proved. Please assert.

Answer 5

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