Since ABCD is a square, there are two things that you know already:

1) The angles DAB, ABC, BCD, CDA are all equal, and they’re all right angles (ninety degrees).

2) The lines AB, BC, CD, DA, are all equal in length.

Now if AP = BQ = CR = DS, then PB = QC = RD = SA

So – given that two triangles with comparable sides and the angle contained between these two sides are equal (in this case, they’re all right angles), that means that the following triangles are all congruent:

SAP is congruent to PBQ is congruent to QCR is congruent to RDS.

This means that the angles ASP, BPQ, CQR, and DRS are all equal. This also means that the angles APS, BQP, CRQ, and DSR are all equal.

Now, since all of these triangles are also right triangles (one angle is 90 degrees, a right angle), and since the angles in any triangle must sum to 180 degrees, this means that the other two angles in each of these four triangles must sum to 90 degrees:

Angles ASP + APS = BPQ + BQP = CQR + CRQ = DRS + DSR = 90 degrees.

But (look back at the congruent triangles) this also means that for the angles

APS + BPQ = BQP + CQR = CRQ + DRS = DSR + ASP = 90 degrees.

Given that the total of all angles at a point on a straight line is 180 degrees, that means that all of the following angles are 90 degrees (or, that they are right angles):

PQR = QRS = RSP = SPQ = 90 degrees.

Now, remember, that all of those triangles were congruent, so the following sides are all equal:

PQ = QR = RS = SP.

So, with all the sides equal, and all the angles being 90 degrees, PQRS is a square.

This is a classic!!! This is one way to prove the pythagorean theorem.

Here is a link for you…

http://www.cut-the-knot.org/pythagoras/index.shtml

Scroll down to Proof number four, and the picture is the same, just with different labeling.

Proof #4

The fourth approach starts with the same four triangles, except that, this time, they combine to form a square with the side (a + b) and a hole with the side c. We can compute the area of the big square in two ways. Thus

(a + b)2 = 4·ab/2 + c2

simplifying which we get the needed identity.

The picture from your problem is almost always used as a teaching tool to “prove” the pythagorean theorem.

Your problem? It has more to do with similar and congruent shapes.

Prove: PB = QC

Given: ABCD is a square

Then, AB, BC, CD, and DA are all congruent (def, of a square)

Given: AP,BQ,CR,DS are congruent

Then:AB=BC

………AB=AP+PB=BC

………AB=AP+PB=BC=QC+BQ

…………….AP+PB=QC+BQ

…………….PB=QC (definition of equality)

Prove: triangle BPQ is congruent to triangle CQR

Given: ABCD is a square

Then Angle A is congruent to Angle C by definition of a square.

PB is congruent to QC (by previous proof)

BQ is congruent to CR (given)

Triangle BPQ is congruent to triangle CQR by SAS.

I will leave step three up to you to write up formally.

My hint is to use angle arguements.

Argue that PQR must be right, because as angle C is a right angle (part of a square), then angles QRC and CQR must be complementary (add to ninety degrees).

And if that is true, using the second proof, you can say that angle PQB is congruent to QRC.

And if THAT is true, then angles PBQ and CQR are also complementary.

AND… if THAT is true, then as segment BQC is straight, and has and angle of 180 degrees, and if PBQ+CQR add to be ninety, then the remaining space created by PQR must be a ninety degree angle.

By a similar method, we can state that angles P,Q,R, and S are all right angles, giving us another square.

I don’t get this: b) Write down 2 reasons to prove that PQRS is a square.

reasons? Maybe the teacher means “ways”. You already have one. Four right angles. The other would be to note that any four sided polygon with four congruent angles will have four congruent sides (def of a square) and you would get those four sides by your step two above by showing that the four small triangles were congruent, so their longest sides are congruent by CPCTC, and they also serve as the sides of the polygon, so it has to be a square.

a)

i)

cause ABCD is a square,so AB=BC=CD=DA,

cause AB=AP+PB,BC=BQ+QC,

so AP+PB=BQ+QC,

cause AP=BQ, so PB = QC.

ii)

cause ABCD is a square,so PBQ=QCR=90(right angle)

from i) PB = QC,

from the question, BQ=CR,

so triangle BPQ is congruent to triangle CQR.

iii)

from ii) ,triangle BPQ is congruent to triangle CQR

so angle BQP=angle CRQ, angle BPQ=angle CQR,

so angle BQP+angle CQR=angle BPQ+angle CRQ,

cause angle BQP+angle BPQ= 180-anglePBQ=180-90=90,

so angleBQP+angle CQR=90.

cause angleBQP+angle CQR+angle RQP=180

so angle RQP=180-angleBQP+angle CQR=90.

solid assertion. The extra ordinary the form, the extra advantageous tasting. I used to shrink my triangles so as that they could all tournament – 2 cuts, nook to nook. Then, sometime somebody confirmed me the thank you to shrink them so as that each and every bit is distinctive. i’ve got by no capacity appeared lower back because. (1st shrink from a million o’clock to eight o’clock positions, 2nd shrink from 10 o’clock to 4.30 positions) possibly one explanation for all it rather is that organic meals infrequently are available one among those uniform commonplace shape, and so our brains react on the sight of the superbly sq. sandwich with horror and apprehension. by way of fact the triangle is far less suitable, (being of countless variability – squares consistently have equivalent aspects) the innovations is extra comfortable, and the entire ingesting adventure is better, manifesting itself as a feeling of extra suitable flavour. It stands to reason that the extra ordinary the form will become, the extra advantageous they’re going to style.

If I don’t put tri before three letters, its an angle.

CPCTC=Congruent Parts of Congruent Triangles are Congruent.

SAS=Side-Angle-Side congruency

There is |||| between step and reason

1) AB=BC |||| def of square

2)AP+PB=AB |||| seg. add. postulate

3)BQ+CQ=BC |||| ” ” ”

4)AP+PB=BQ+CQ |||| substitution of 1 with 2 and 3

5)AP=BQ |||| given

6)BQ+PB=BQ+CQ |||| substitution of 4 with 5

7)PB=CQ |||| subtraction

8)ABC=90 |||| def square

9)BCD=90 |||| ” ”

10)ABC=BCD |||| substitution of 8 with 9

11)BQ=RC |||| Given

12)tri PBQ=tri CQR |||| SAS with 10,11,7

13)PQB=QRC |||| CPCTC

14)PBQ+BQP+QPB=180 |||| 3 angles in tri add to 180

15)90+BQP+QPB=180 |||| Substitution of 14 with 8

16)BQP+QPB=90 |||| Subtraction

17)BQP+QRC=90 |||| substitution of 16 with 13

Create X on PQ where P is between X and Q

Create Y on RQ where Q is between Y and R

18)PRY=QRC |||| Vertical Angles

19)XPA=BPQ |||| Vertical Angles

20)PRY+XPA=90 |||| Substitution of 17 with 18 and 19

21)PQR=90 |||| Angle in tri equals sum of alternate exterior angles

I’m not going to do your homework for you, but you can prove congruent triangles using the ASA, SAS, SAA, or HL theorems. If you don’t know what those are, please try to look them up before asking others.

all i can tell you is to substitute in numbers for the sides of both squares and go from there it should be a lot easier.