Then from the AADB, since DE is drawn perp. to AB,

BD2 = AD2 + AB2 - 2AB. AE [II. 13].

Again from the ▲ADB, since BC is drawn perp. to AD produced, BD3 = AD2 + AB2 - 2AD. AC [II. 13].

.. AB. AE = AD. AC.

25. Let ABCD be the par"; and let the diags. meet at O. Then O is the middle point of AC and BD [p. 64, Ex. 5]. Then by the last Ex., from ▲ ABC, AB2 + BC2 = 2A02 + 20B2. Also, from ACDA, CD2 + DA2 = 20c2 + 20D2.

=

By addition, remembering that AO OC and OB = OD,

AB2 + BC2 + CD2 + DA2 = 4A02 + 40B2

=AC2 + BD2 [Ex. 1, p. 144].

26. Let ABCD be the quad., and P, Q, R, S, the middle points of the sides AB, BC, CD, DA.

Then PQRS is a parm. [Ex. 9, p. 97]

and AC = 2PQ, also BD = 2SP [Ex. 3, p. 97].

.. AC2+ BD2 = 4PQ2 + 4SP2

27.

Let ABCD be the rect., and P the given point within it. Let the diagonals meet at O. Then AC = BD [1. 4] and the diags. bisect one another [Ex. 5, p. 64]. .. AO DO.

Then from

and from

28.

BD, AC.

APC, PA2+ PC2 = 2 [AO + PO2] [Ex. 24, p. 147], ▲ BPD, PB2 + PD2 = 2 [DO2 + PO3],

Let ABCD be the quad., and X, Y the middle points of
Join YB, YX, YD.

Then from the ABC, AB2 + BC2 = 2 AY2 + 2BY2 [Ex. 24, p. 147]; also from the ▲ ADC, AD2 + DC2 = 2AY2 + 2DY2.

.. AB2 + BC2 + AD2 + DC2 = 4AY2 + 2 (BY2 + DY2)

= AC2 + 4 (DX2 + XY2) [Ex. 24, p. 147]

= AC2 + BD2 + 4XY2.

=

29. From the APB, AP2 + BP2 = 2A02 + 20p2 [Ex. 24, p. 147].

But both AO and OP are constant,

.. AP2 + BP2 is constant.

30. Let BC be the base, and ABC the A in one of its positions.

Bisect BC at X, and join AX.

Then BA2 + AC2 = 2 (BX2 + AX2) [Ex. 24, p. 147].

But BA2 + AC2 is constant by hyp.

.. BX2 + AX2 is constant.

And since BX is constant, .. AX is constant; and X is a fixed point; .. the locus of A is a circle, whose centre is at X.

31. Since CB is a median of the AACD,

But

.. DC2 + CA2 = 2AB2 + 2BC2 [Ex. 24, p. 147].

32. Let ABC be a ▲ rt. angled at B, and let the hypotenuse be divided into three equal parts AE, EF, FC.

Then from the

and from the

ABF, AB2 + BF2 = 2BE2 + 2EF2 [Ex. 24, p. 147] ▲ EBC, BE2 + BC2 = 2BF2 + 2EF2.

Hence by addition

-

33. Let AX, BY, CZ be the medians of the ABC.

Then

and

also

AB2 + AC2 = 2AX2 + 2BX2 [Ex. 24, p. 147],

AB2 + BC2 = 2BY2 + 2AY2,

BC2+ AC2 = 2cz2 + 2AZ2.

By addition

2AB2 + 2BC2 + 2AC2 = 2 (AX2 + BY2 + Cz2) + 2BX2 + 2AY2 + 2AZ2 and the doubles of these equals are equal, so that

4AB2 + 4BC2 + 4AC2 = 4 (AX2 + BY2 + Cz2) + 4BX2 + 4AY2 + 4AZ2 = 4 (AX2 + BY2 + CZ2) + BC2 + AC2 + AB2.

Hence 3 [AB2 + BC2 + AC2] = 4 [AX2 + BY2 + CZ2].

34. Let AX, BY, CZ be the medians, intersecting at O. OA=20X, OB = 20Y, OC=2oz [Ex. 4, p. 105], A BOC, OB2 + OC2 = 2BX2 + 20x2.

Then

and from the

20A2 + 20B2 + 20c2 = 2Bx2 + 2cY2 + 2Az2 + 20x2 + 20y2 + 2oz2, and doubling these equals, we have

40A2 + 40B2 + 40C2 = 4BX2 + 4CY2 + 4AZ2 + 40x2 + 40Y2 + 4oz2 = BC2 + CA2 + AB2 + OA2 + OB2 + OC2.

and

Hence

3 {OA2 + OB2 + OC2} = BC2 + CA2 + AB2.

35. Let H and K be the middle points of the diags. BD, AC. Now

By addition

PA2 + PC2 = 2AK2 + 2PK2 [Ex. 24, p. 147],

PB2 + PD2 = 2BH2 + 2PH2.

PA2 + PB2 + PC2 + PD2 = 2AK2 + 2BH2 + 4XH2 + 4XP2

= 2AK2 + 2XK2 + 2BH2 + 2XH2 + 4XP2

= XA2 + XC2 + XB2 + XD2 + 4XP2.

36. Let ABCD be the trapezium, AB and DC being the par1. sides, and let DC be less than AB.

also

Draw CX, DY perp. to AB.

Then AC2 = BC2 + AB2 - 2AB. BX [II. 13],

BD2 = AD2 + AB2 - 2AB. AY;

.. AC2 + BD2 = BC2 + AD2 + 2AB2-2AB. BX - 2AB. AY.

But AB2 = AB. BX + AB. XY + AB. AY.

[II. 1.]

So that 2AB2 - 2AB. BX - 2AB. AY 2AB. XY = 2AB. CD.

Hence

AC2 + BD2 BC2 + AD2 + 2AB. CD.

37. greater.

PROBLEMS.

Let H and K be sides of the given sqq., of which H is the

Draw AP equal to H; produce AP to B making PB equal to H; and from PB cut off PQ equal to K.

Then the rect. AQ, QB is that required.

For since AB is divided equally at P and unequally at Q, .. AQ. QB PB2- PQ2 [11.5]

38. Let BF be the st. line, and K a side of the given sq. [See fig. p. 143.]

On BF describe a semi-, and from any point X in BF, or BF produced, draw XY perp. to BF, making XY equal to K. Through Y draw YHH' par. to BF cutting the semi- at H and H'. From H (or H') draw HE perp. to BF.

sq.

Then shall BE. EF = HE2 = K2. [Proof as in 11. 14.]

Then

39. Let BE be the side of the rect., and K a side of the given Join At E draw EH perp. to BE, making EH equal to K. BH; and draw HF perp. to BH to meet BE produced at F. shall EF be the other side of the rect. For [p. 59, Ex. 4, or III. 31] a semicircle described on BF will pass through H.

Hence BE. EF = EH2 = K2. [Proof as in II. 14.]

40. Let AB and X be the two given st. lines.

Analysis. Let AB be bisected at P and produced to Q.

Thus the length of PQ may be found by drawing a rt.-angled triangle [1. 47]. And as P is a fixed point, Q is determined.

41. This is the same as dividing a line externally in medial section. [See Ex. 21, p. 146, and p. 139, note.]

42. Draw the rect. ABDH, contained by AB and X.

Take from these equals the common fig. AK.

Then fig. FC = fig. CD.

Or AC2 BC. X.

=

BOOK III.

EXERCISES.

Page 156.

1. The st. line bisecting AB at rt. ▲ is the st. line passing through both centres.

2. Let the bisector of BAC cut BC in D. ADB, ADC, it follows that DB = DC and ADB = .. AD passes through the centre.

Then in As

ADC [1. 4].

3. If the chords are not par1., bisect each at rt. s. If they are par1., join their extremities: and bisect each of these new chords at rt. s. The pt. of intersection of the bisectors is the required centre.

B, C be the three given
Join OB, OC. Then in
OCY, OAY, OC = OA.

4. See fig. to Ex. 1, p. 103. Let A, pts. Let XO, YO bisect BC, AC at rt. 4. A3 OBX, OCX, OC = OB [1. 4] and in A .. OA = OB = OC, and O is the centre of the required O.

8

5. The required locus is the st. line bisecting at rt. the st. line joining the two given pts.

6. With the two given pts. as centres, describe circles each of which has a radius equal to the given radius. Again, with the pt. of intersection of these circles as centre, describe a circle with the given radius. This is the circle required.