4. Let BD be the diagonal of given sq. ABCD, then sq. on DB = sum of sqq. on DC, BC

5. In ▲ ABX, ACX,

LABXL ACX,

Because AXB = AXC, being rt. 2o. and AX is common;

..▲ ABX, ACX are equal in all respects:

... BX = XC; or BC is twice BX.

... sq. on BC= 4 times sq. on BX,

that is, sq. on AD = 4 times sq. on BX.

But sq. on AB = sum of sqq. on BX, AX;

.. sq. on AX=3 times sq. on BX.

[1. 26.]

6. Let AB, CD be sides of the two squares; at B draw BE Join AE. perp. to AB and equal to CD.

Then sq. on AE = = sum of sqq. on AB, BE

= sum of sqq. on AB, CD.

7. Sq. on AB sum of sqq. on AX, BX,

sq. on AC = sum of sqq. on AX, XC;

.. diff. of sqq. on AB, AC = diff. of sqq. on BX, XC.

8. By Ex. 7,

sq. on BZ - sq. on AZ = sq. on OB-sq. on OA,
sq. on AY - sq. on CY=sq. on OA- sq. on OC,
sq. on CX-sq. on BX = sq. on OC -- sq. on OB;
.., by addition,

sum of sqq. on BZ, AY, CX - sum of sqq. on AZ, CY, BX = 0, or, sum of sqq. on BZ, AY, CX= sum of sqq. on AZ, CY, BX.

.. sum of sqq. on BQ, PC = sum of sqq. on AB, AC, together

with sum of sq. on AQ, AP,

=sum of sqq. on BC, PQ.

10. Let the medians be BQ, CP; then it may be shewn that BC is double of PQ. [See solution of Ex. 3, p. 105.]

.. sq. on BC= 4 times sq. on PQ.

And by Ex. 9,

sum of sqq. on BQ, CP= sum of sqq. on BC, PQ

3. Let the base BC be bisected by AD, which also bisects the

vert. BAC.

Produce AD to E, making DE equal to DA, and join CE.

4. Let BO, CO, drawn from extremities of the base, meet at O.

BC.

5. Let BD, CE be the equal perps. drawn from ends of base

7. Let BX, CY be drawn perp. to AC, AB respectively, and let them meet in O. Join AO.

Then ▲ BAO, CAO have their sides respectively equal;

..A AKB is isosceles [1. 6]. Similarly A KDC is isosceles.

11. Here the greatest angle is a right angle.
Hence the required result follows by Ex. 4 on p. 65.

[1. 32.]

7. Let O be the given pt., AC and BD the diagonals; then AO + OC > AC, BO + OD > BD [1. 20]. The exceptional case is when O is at the intersection of the diagonals.

8. Let the median AD bisect BC; produce AD to E making DE equal to AD, join EC. Then A ABD, EDC are identically equal [1. 4] and AB = CE. Now AC + CE > AE [i. 20]. That is, AB+ AC > 2AD.

9. This follows at once from Ex. 8, since twice the sum of the sides is greater than twice the sum of the medians.

10. Let the median AD bisect BC. If AD > DC, LACD is greater than ▲ DAC; similarly DBA is greater than DAB. Hence the sum of the angles at B and C is greater than the angle at A; that is, BAC is acute [1. 32]. The other cases follow similarly.

11. In the rhombus ABCD let DAB be greater than ABC. Then since the sides of a rhombus are equal it follows that

Thus by Ex. 12, AP lies between AD and AX, and by Ex. 3 it is intermediate between them in magnitude.

III. ON PARALLELS. Page 103.

2. From O any pt. on the bisector of ▲ BAC draw OP par1. to AB, and OQ par1. to AC. Then QOA LOAP = LOAQ.

.. QO = AQ=OP since OPAQ is a parm. Also OQ=AP; thus the fig. is equilat.