For a geometry problem, 50 points will be added if you get it right

Let the intersection of AD and BC be O, and the inner angle of the triangle is 180 degrees, and the angle CDB=60 degrees, that is, X+Y=60, the angle BAD=180-40-70-Y=70-Y, and the angle AOB=BCD+CDA=50+X, so in the triangle AOB, 40+(50+X)+(70-Y)=180, that is, X-Y=20, the solution is X=40, Y=20

Solution: (1) The quadrilateral ABCD is rectangular (known).

da=bc cd=ab(rectangular sides are equal) cda= abc (all four corners of the rectangle are right angles) acd is all equal to cab(sas).

cab = acd (congruent triangles correspond to equal sides) edc = cab (known).

ACD = EDC (Equal Substitution).

ac de (equal internal wrong angles, two straight lines parallel).

2) dec=90°,bf ac(known) afb= dec=90°,edc= cab,ab=cd(known) dec afb,de=af,by ac de,quadrilateral afed is a parallelogram, ad ef and ad=ef, in rectangular abcd, ad bc and ad=bc, ef bc and ef=bc, quadrilateral bcef is a parallelogram

In rectangular ABCD, ac de, dca= cab, edc= cab, dca= edc, ac de;

The quadrilateral BCEF is a parallelogram

Reason: Afb= Dec=90° is obtained from dec=90°,bf ac, and edc= cab,ab=cd, dec afb, de=af,ac de, from ac de, the quadrilateral afed is a parallelogram, ad ef and ad=ef, in the rectangular abcd, ad bc and ad=bc, ef bc and ef=bc, and the quadrilateral bcef is a parallelogram

(1) Proof:

Since ABCD is rectangular, ADC= BAD=90 degrees, and because EDC= CAB

DAC + CAB = 90 degrees.

So: EDC + ADC + DAC

cab+90+90-∠cab

180 degrees. So: ac de (complementary to the side inner angle, two straight lines parallel) (2) cause, afb = dec = 90°

ab=dcedc=∠cab

So the triangle EDC is all equal to the triangle FAB

So: ec=bf

ecd=∠fba

Because ECB+ CBF= ECD+90+90- FBA=180 degrees.

So EC fb

So: the quadrilateral BCEF is parallel and equal to the opposite side EC, FB, and is a parallelogram.

1.Because ABCD is rectangular.

So dc ab

Because edc= cab

So ac de

2.Because the triangle ABF is all equal to the triangle DCE

So bf=ce, af=de

So ad=ef

So bc=ef

So the quadrilateral bcef is a parallelogram.

in rectangular ABCDs;

1、∵dc∥ab

dca=∠cab

edc=∠cab

de∥ac1、∵∠dec=90º,de∥ac

ecf=∠cfb=90º

CE BF is also dec= afb, dc=ab edc= cab edc fab

ce fbbcef is a parallelogram.

As shown in the figure, as p r1 oa, pq1 ob, and the vertical foot is r1 and q1, it is easy to prove that rt pr1r congruent rt pq1qpr=pq

So the conclusion is that when rotating, PR is equal to PQ.