(This is proved in the previous deduction when the point is taken within the figure, and it is proved in the same way when the point is taken without the figure.)

B

9. Prove that the sum of the distances of any point from the angular points of a quadrilateral is greater than half the perimeter of the quadrilateral.

Let ABCD be a quadrilateral, and let P be any point, either within or without the quadrilateral ABCD.

Then the sum of the distances of P from the angular points of ABCD shall be greater than half the perimeter of ABCD.

B

Draw PA, PB, PC, PD (post. 1).

Proof-The two sides PA, PB are greater than the side AB (I. 20);

the two sides PB, PC are greater than the side BC;

the two sides PC, PD are greater than the side CD;

and the two sides PD, PA are greater than the side DA; therefore twice PA, PB, PC, PĎ are together greater than AB, BC CD, DA.

Wherefore the sum of the distances of P from the angular points of ABCD is greater than half the perimeter of ABCD. Q.E.D.

10. The diagonals of a rhombus bisect each other at right angles.

Let ABCD be a rhombus, of which AC, BD are the diagonals.

Then AC, BD shall bisect each other at right angles.

B

Let O be the intersection of the diagonals.

Proof. In the triangles BAC, DAC,

the two sides BA, AC are equal to the two sides DA, AC, each to each,

and the base is equal to the base DC;

therefore the angle BAC is equal to the angle DAC (1. 8). Again, in the triangles BAO, DAO,

the two sides BA, AO are equal to the two sides DA, AO, each to each,

and the angle BAO has been proved to be equal to the angle

DAO;

therefore the base BO is equal to the base DO (1. 4),

and the angle BOA is equal to the angle DOA; and these are adjacent angles;

therefore AO is at right angles to BD (def. 10): wherefore AC, BD are at right angles.

It has just been proved that BO is equal to OD;

similarly it can be proved that AO is equal to OC:

wherefore AC, BD bisect each other.

Therefore AC, BD bisect each other at right angles. Q.E.D.

11. Describe a square which shall be equal to the difference between two given squares.

Let A, B be the sides of the two given squares.

It is required to describe a square which shall be equal to the difference of the squares on A, B.

D

Draw the straight line CD equal to A, the side of the larger square;

from the centre D, at the distance DC, describe the circle CEF (post. 3);

produce CD to meet the circumference in E (post. 2);

from DE cut off DG equal to B, the side of the smaller square

(1. 3);

at G draw GF at right angles to CE (1. 11), meeting the cir

cumference in F.

Then GF shall be the side of the required square.
Join DF.

Proof. The square on DF is equal to the square on DG, GF (L. 47);

but DF is equal to DC (def. 15), and therefore equal to A (ax. 1);

and DG is equal to B (con.);

therefore the square on A is equal to the squares on B, GF: wherefore the square on GF is equal to the difference of the squares on A, B (ax. 3). Q.E.F

12. Find a point in the diagonal of a square produced, from which, if a straight line be drawn parallel to any side of the square, and meeting another side produced, it will form, together with the produced diagonal and produced side, a triangle equal to the square.

Let ABCD be a square, of which AC is a diagonal.

It is required to find a point in AC produced, such that the triangle formed by AC produced to this point, by the line drawn parallel to a side of the square through the point, and by the side of the square produced to meet this parallel line, shall be equal to the square ABCD.

B

E

Produce AB to E, making AE equal to the diagonal AC (1. 3); through E draw EF parallel to BC (1. 31), and meeting AC produced in F.

Then F shall be the point required, that is, the triangle AEF shall be equal to the square ABCD.

Proof.-Because BA is equal to BC (hyp.),

the angle BAC is equal to the angle BCA (1. 5); but the angle EFA is equal to the angle BCA (1. 29); therefore the angle EAF is equal to the angle EFA (ax. 1): wherefore the side EA is equal to the side EF (1. 6).

The rectangle FE, EA is double of the triangle AEF (1. 41); but the rectangle FE, EA is equal to the square on AE, that is, equal to the square on the diagonal AC (con.); therefore the square on AC is double of the triangle AEF. Again, the square on AC is equal to the squares on AB, BC (I. 47),

therefore the square on AC is double of the square ABCD; but it has been proved that the square on AC is double of the triangle AEF;

wherefore the triangle AEF is equal to the square ABCD (ax. 7). Q.E.F.

13. The quadrilateral figure whose diameters bisect each other is a parallelogram.

Let ABCD be a quadrilateral, and let the diameters AC, BD bisect each other in E.

Then ABCD shall be a parallelogram.

B

Proof-In the triangles AED, BEC,

the two sides AE, ED are equal to the two sides CE, EB, each to each (hyp.),

and the angle AED is equal to the angle CEB (1. 15); therefore the base AD is equal to the base CB (1. 4), and the angle ADE is equal to the angle CBE.

Because the angle ADE is equal to the alternate angle CBE, therefore AD is parallel to BC (1. 27):

wherefore AD, BC are both equal and parallel; therefore AB, CD are both equal and parallel (1. 33). Wherefore ABCD is a parallelogram (def. A.). Q.E.D.