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

BDP = AD? + AB– 2AB. AE (11. 13].
Again from the AADB, since BC is drawn perp. to AD produced,

BD* = ADP + AB– 2AD .AC (11. 13].

.. AB. AE = AD. AC.

25. Let ABCD be the par"; and let the diag®. meet at o.
Then o is the middle point of AC and BD (p. 64, Ex. 5].
Then by the last Ex., from A ABC, AB® + BC = 2A02 +20B?.
Also, from ACDA, CD? + DA? = 2002 + 2002.
By addition, remembering that AO = OC and OB=OD,
AB? + BC? + CD+ DAP = 4A02 + 40B2

= AC? + BD? [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 par". [Ex. 9, p. 97]

and AC = 2PQ, also BD = 2SP [Ex. 3, p. 97]. .. AC? + BD2 = 4PQ2 + 4sp2

= 2 {PQ+ SR? + SP2 + QR?}
= 2 {PR? + SQ?} [Ex. 25, p. 147]-

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 diag®. bisect one another [Ex. 5, p. 64]. .. AO = DO.

Then from A APC, PA + PC? = 2 [AO? + PO?] [Ex. 24, p. 147], and from A BPD, PB? + PDP = 2 [DO? + PO®],

:. PA? + PCP = PB? + PDP.

28. Let ABCD be the quad., and X, Y the middle points of BD, AC.

Join YB, YX, YD. Then from the AABC, AB? + BCP = 2 AY2 + 2BY? [Ex. 24, p. 147]; also from the A ADC, ADP + DC2 = 2AY2 + 2DY?. :. AB + BC? + ADP + DC= 4AY2 + 2 (BY? + DY2)

= AC + 4 (DX2 + XY) [Ex. 24, p. 147] : AC+ BD? + 4XY?.

29. From the AAPB, AP? + BP= 2A0+ 20P2 [Ex. 24, p. 147]. But both AO and OP are constant, .. AP? + BP2 is constant.

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

Bisect BC at X, and join AX.
Then BA’ + AC= 2 (BX® + Ax2) [Ex. 24, p. 147].
But BA? + ACis constant by hyp.
.. BX? + 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,

:: DC? + CAP = 2AB” + 2BC? [Ex. 24, p. 147]. But

CAP = AB? (Hyp.] .. DC2 = AB? + 2BC?.

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

Then from the A ABF, AB? + BF? = 2BE2 + 2EF? [Ex. 24, p. 147]
and from the AEBC, BE? + BC? = 2BF2 + 2EF?.
Hence by addition
AB? + BC+ BF? + BE? = 2BF? + 2BER + 4EF,

ACP = BFP + BE? + 4EF.
But, since AC = 3EF, .. AC= 9EF?

:. 5EF2 = BF? + BE?.

or,

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

AB? + AC? = 2AX + 2BX? [Ex. 24, p. 147], and

AB? + BC= 2BY+ 2AY?, also

BC? + AC? = 2cZ2 + 2AZ?.

By addition 2AB+ 2BC? + 2AC2 = 2 (AX2 + BY2 + CZ2) + 2BX2 + 2AY? + 2AZ? and the doubles of these equals are equal, so that 4AB2 + 4BC2 + 4AC? = 4 (AX2 + BY? + CZ2) + 4BX2 + 4AY2 + 4AZ?

= 4 (AX? + BY2 + CZ2) + BC? + AC? + AB?.
Hence 3[AB+ BC+ AC*] = 4[AX? + BY® + CZ].
34. Let AX, BY, CZ be the medians, intersecting at o.

Then OA = 20%, OB = 20Y, OC = 202 [Ex. 4, p. 105], and from the A BOC, OB? + Oc?= 2BX2 + 20x2. Again from

ACOA, oc? + OA? = 2CY? + 20Y, also from A AOB, OA? + OB? = 2AZ? + 2022.

:. by addition 20A’ + 20B2 + 2002 = 2BX2 + 2CY2 + 2AZ2 + 20x2 + 20y2 + 2022, and doubling these equals, we have 40A? + 40B + 400? = 4BX2 + 4CY? + 4AZ+ 40x2 + 40Y2 + 4022

=BO2 + CAP + AB? + OA? + OB? + Oc?. Hence 3 {OA? + OB? + OC?} = BC2+ CA? + AB”. 35. Let H and K be the middle points of the diags. BD, AC. Now

PAR + PC?=2AK2 + 2PK? (Ex. 24, p. 147), and

PB? + PD2= 2BH2 + 2PH?.
By addition
PA? + PB? + PC? + PD2 = 2AK? + 2BHP + 4XH2 + 4Xp2

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

XA2xco +XB +XD2 + 4XP?. 36. Let ABCD be the trapezium, AB and DC being the par. sides, and let DC be less than AB.

Draw CX, DY perp. to AB.

Then AC? = BC? + AB? – 2AB. BX (11. 13], also BD2 = AD? + AB2 – 2AB. AY; .: AC + BD? = BC? + ADP + 2AB2 — 2AB. BX – 2AB. AY. But ABP = AB. BX + AB. XY + AB. AY. [11. 1.] So that 2AB2 – 2AB. BX - 2AB. AY = 2AB. XY = 2AB. CD. Hence AC? + BD2 = BC? + AD2 + 2AB. CD.

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PROBLEMS. 37. Let H and K be sides of the given sqq., of which H is the greater.

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 = PB? – PQ? (11. 5]

= sq. on H - sq. on K.

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

On BF describe a semi-o, 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-o at H and H'. From H (or H') draw HE perp. to BF.

Then shall BE. EF = HE? = K?. [Proof as in 11. 14.]

39. Let BE be the side of the rect., and K a side of the given sq. At E draw EH perp. to BE, making EH equal to K. Join BH; and draw HF perp. to BH to meet BE produced at F. Then shall EF be the other side of the rect. For [p. 59, Ex. 4, or 111. 31] a semicircle described on BF will pass through H.

Hence BE. EF = EHP = K?. [Proof as in 11. 14.]

40. Let AB and X be the two given st. lines. Analysis. Let AB be bisected at P and produced to Q. Then

AQ. QB = PQ? – PB?. [11. 6.]
Required

AQ. QB = X
Hence we must have PQ2=X2 + PB2.

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.
Produce HA to F, so that

HF.FA= AB. X (11. 14 and Ex. 40, p. 148].
On FA describe a sq. AFGC as in 11. 11. Produce GC to meet
HD at K.

Then fig. FK = fig. AD,
Take from these equals the common fig. AK.
Then fig. FC = fig. CD.
Or AC2= BC. X.

BOOK III.

EXERCISES.

If they

Page 156. 1. The st. line bisecting AB at rt. _ s is the st. line passing through both centres.

2. Let the bisector of <BAC cut BC in D. Then in As ADB, ADC, it follows that DB = DC and LADB = L ADC [1. 4). .. AD passes through the centre.

3. If the chords are not par?., bisect each at rt. 2 are par., join their extremities : and bisect each of these new chords at rt. Ls. The pt. of intersection of the bisectors is the required centre.

4. See fig. to Ex. 1, p. 103. Let A, B, C be the three given pts. Let XO, YO bisect BC, AC at rt. 28 Join OB, OC. Then in AS OBX, OCX, OC = OB [1. 4) and in As OCY, OAY, OC = OA. .. OA = OB = OC, and o is the centre of the required O.

5. The required locus is the st. line bisecting at rt. 48 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.

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