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7. The ZACB of a A ABC is four-thirds of a right angle; show that AB2= AC2+BC2+AC BC.
8. The ZACB of a A ABC is two-thirds of a right angle; show that AB2= AC2+BC2 - AC.BC.
Use $ 13, Ex. 1, and Euc. II. 13. 9. If, in the A ABC, AC=BC, and < ACB=four-thirds of a right angle; show that AB2=3AC2. 10. O is a point within the A ABC such that
_ BOC= _COA= LAOB; show that BC2+CA2+AB2=2(0A2+OB2+0C2)+OB:OC+OCOA+OA.OB. 11. If, in the A ABC, BQ and CR are the altitudes from E and C, and the 28 B, C are both acute; show that
Use Euc. II, 12 or 13, and Ex. (1). Consider the case in which the < B is obtuse.
12. H is the orthocentre of the acute-angled A ABC and AP, BQ, CR are the altitudes ; show that
(2) AH+BCP=BHP+CA2=CH2+AB?. For (1) use Ex. 1, for (2) use § 20, Ex. 5 or Euc. II. 12.
13. In the A ABC, AC=BC, and AN I BC, or BC produced ; show that AB=2BC-BN.
Projections. 14. If two straight lines be equal and parallel, their projections on any other straight line shall be equal.
Enunciate and prove two converses to this theorem. 15. A, B are points on the bisector of the < XOY; P, Q are their projections on OX; and R, S are their projections on OY; show that PR QS.
NOTE. — The projection of a point on a straight line is the foot of the perpendicular drawn from the point to the line.
$ 39. Problems which follow directly from known propositions.
1. Divide a given straight line into two parts so that the rectangle contained by the parts shall be equal to a given square.
This is a converse problem to Euc. II. 14, and the following construction is at once suggested.
Bisect AB in O, and describe a circle with O as centre and OA as radius.
Through o draw OCIAB and equal to a side of the given square.
Through C draw CP | AB, meeting the circle in P, and through P draw PQLAB.
Q shall be the required point of section.
[Euc. II. 5. =PO2;
=PQ2+0Q2; [Euc. I. 47. ..AQ.QB=PQ2.
Observe that if PC be produced to meet the circle again in P', and P'Q' be drawn I AB, a second point will be found to satisfy the condition, but that, if OC >OA, no point satisfies the condition.
Observe, also, that APB is a right-angled triangle, and compare § 32, Ex. 1.
2. Construct a square equal to n times a given square.
Let AB be a side of the given square.
and proceed as in Euc. II. 14. This problem may also be solved by means of Euc. I. 47.
3. Construct a rectangle equal to a given square and having one side equal to a given straight line.
Let BC be a side of the given
Use $ 32, Ex. 1.
4. Produce a given straight line, so that the rectangle contained by the whole line thus produced, and the part produced, shall be equal to a given square.
Let AB be the given straight line, CB a side of the given square.
Produce AB and make OP=OC where O is mid-point of AB.
Use Euc. II. 6.
5. Produce a given straight line AB to P, so that ABAP shall be equal to the square on a given straight line AC.
See $ 32, Ex. 1 (2).
6. A, B are fixed points and XY is a given straight line; find a point P in XY, such that AP2+BP2 may be a minimum.
Bisect AB in O, and draw OPIXY.
Use $ 37, Ex. 1, and $ 8, Ex. 1. 7. Construct a rectangle having given the area and the perimeter.
Observe that in Ex. 1, AB is one-half the perimeter. 8. Construct a square which shall be equal to
(1) A given parallelogram.
(2) A given triangle. 9. Construct a rectangle equal to the sum of two given squares, so that one side of the rectangle may be equal to a given straight line.
10. Construct a rectangle equal to the difference of two given squares, so that one side of the rectangle may be equal to a given straight line. 11. Divide AB in P, so that
(1) AB2+PB2=3 AP2.
(2) AP2 - PB2= AP-PB. See $ 36, Ex. 2. Can P be a point of external section ?
12. Produce AB to P so that AB:BP may be equal to a given square.
See $ 32, Ex. 1. 13. Divide a straight line so that the rectangle contained by the segments may be a maximum.
Use Euc. II. 5. 14. Divide a straight line so that the sum of the squares on the segments may be a minimum.
Use Euc. II. 9.
$ 40. Analysis and Synthesis.
1. Divide one side of a triangle, so that the sum of the squares on its segments may be equal to the sum of the squares on the other two sides of the triangle.
AB2+ ACP= PB2+PC.
AB2+AC?=2AD2+2 BD2. [S 37, Ex. 1.
Bisect BC in D.
P is the point required.
[S 37, Ex. 1. =2PD2+2BD
[Construction. PB2+ PC
[Euc. II. 9, 10. A second solution can be obtained by cutting off
DP'=AD Observe that P is in BC, or BC produced, according as LA is obtuse or acute, and falls on B when _ A is right. Compare $ 12, Ex. 6.