PROPOSITION XXXIV.

The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.

Let FGHK be a, and GK one of its diagonals, and SR, PN □s about the diagonal.

Then FQ, QH, the remainings, which are called complements, shall be equal to one another.

Fors are bisected by their diagonals, (1. 26) .. ASGQ is = ARGQ, and ▲ PQK = ^ NQK;

.. AS SGQ, PQK are together

but the whole ▲ FGK is

=

=

as RGQ, NQK;

the ▲ HGK;

.. the remainder FQ is the remainder QH.

PROPOSITION XXXV.

In any right-angled triangle, the square described on the side opposite the right angle is equal to the squares described on the sides containing the right angle.

UN

is the right QPH; add to each the RPQ;

=

RPH.

Hence SP, PQ and the 4 SPQ are respectively = RP,

Now the PC is double of the ▲ RPH, because they are on the same base and between the same parallels. (1.32) FRP and QRP are right ▲ s;

Also

L

...FRQ is a straight line (1. 10), and it is || to SP.
SPRF is double of the ▲ SPQ,

..

(I. 32)

.. PC is = SPRF the square on PR. Similarly it may be proved that the QC is = the square on RQ;

is

=

.. the whole figure PHNQ, that is the square on PQ, the squares on PR, RQ.

PROPOSITION XXXVI.

If the square described on one of the sides of a triangle be equal to the squares described on the other two sides of it, the angle contained by those sides is a right angle.

R

Q

Let the square described on the side PQ of the ▲ PQR be equal to the squares on PR, RQ.

Then shall the ▲ PRQ be a right ▲.

Through R draw RS 1 to PR and = RQ.

Join SP.

Now. SR is = QR; .. square on SR is

.. the squares on SR, RP are = the squares on QR, RP, but the squares on SR, RP are together the square on SP;

· SRP is a right 4;

=

(1.35)

and the squares on QR, RP are = the square on QP; (hyp.)

Hence, PR, RS, SP are respectively = PR, RQ, QP.

the square on QP;

PROBLEM L.

Describe a parallelogram equal to a given triangle, and having an angle equal to a given angle.

Let FGH be the given A, and K the given angle. It is required to describe a = A FGH and having an angle = K.

Bisect GH in R; join FR; through F draw FST || to GH; at R in RH make 4 HRS = K.

Through H draw HT || to RS;

then SRHT shall be the required.

... GR is

=

RH;. AFGR is

= ▲ FRH; (1. 32)

. A FGH is double of the & FRH;

also the SH is double of ▲ FRH,. they are upon the same base and between the same parallels.

(1. 32)

..SH is = ▲ FGH, and it has an SRH equal to

the given angle K.

PROBLEM M.

To a given straight line apply a parallelogram equal to a given parallelogram.

Let FG be the given straight line, and LH the given ☐ .
It is required to apply to FG a = LH.

Produce GF, and cut off FM = HK,

and on FM describe PF equal in all respects to LH. Through G draw GQ to PM meeting PC produced in Q. Join QF and produce it to meet PM produced in S. Through S draw SVR || to MG or PQ meeting CF and QG produced in V and R.

Then FR is a, and it shall be the one required.

For FR, PF are the complements about the diagonal QS of the PR, and they are.. = one another,

... FR is

=

(I. 34)

LH and it has been described on FG.

COR. Hence to a given straight line may be applied a parallelogram equal to a given triangle and having an angle equal to a given angle.

=

For a LH may be drawn the given ▲ and having an = the given 4, and the remainder of the construction

is the same as in the proposition.