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11. G is the point of intersection of the medians of the A ABC, and P is any other point; ehow that PA2+PB2+PC2=GA2+GB2+GC2+3PG2.

(A Standard Theorem.)

B

Apply Ex. 1 to A PBC, and Ex. 10 to A PAD, and deduce

that PAP+PB2+PC2=2BD2+6 DG2+3PG?. Use Ex. 1 to show that

GA2+GB2+GC2=2BD2+6DG?.

12. The hypotenuse BC of a right-angled A ABC is trisected in D and E; show that AD2+AE2=5DE2.

13. If two medians of a triangle be equal, the triangle shall be isosceles.

Use Ex. 1.

14. ABCD is a trapezium in which AD is parallel to BC; G, H are the mid-points of AB, CD; show that

AC2+AD2+BC2+BD2=AB2+CD2+4GH2. 15. K is the point of intersection of the diagonals AC and BD of the quadrilateral ABCD, and O is the mid-point of the straight line joining the mid-points of the diagonals; show that KA2+KB2+KCP+KD2=OA2+OB2+002+OD2+4K02.

Use Euc. II. 9 and Ex. 1.

16. ABCD is a diameter of two concentric circles, and P, Q are points on the inner and outer circles respectively; show that PA2+PD2=QB2+QC2.

17. ABC is a triangle, and D is a point in BC, such that 3BD=DC; show that 3AB2+AC2=12BD2+4AD2.

Compare Ex. 10. 18. A, B, C, D are four points, E is the mid-point of AB; EC is divided in F, so that 2EF=FC; FD is divided in G, so that 3FG=GD, and P is any point; show that

PA2+PB2+PC+PD2=GA2+GB2+GC2+GD2+4PG".

$ 38. (Bookwork, EUCLID, II. 1-14.)

1. Two right-angled As ACB, ADB have the same hypotenuse AB; AD, BC intersect in 0; it is required to prove that

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In figure 1, 2 AOB is obtuse.

.. AB2=AO+OB2+2AO OD. Similarly, AB2=AO+OB2+2BO.OC;

.::AO.OD=BO.OC.

[Euc. I. 16. (Euc. II. 12.

In figures 2 and 3, 2AOB is acute.

... AB2=AO+OB2— 2AO.DO. Similarly, AB2=A02+AO2-2BO.CO;

.. AO DO=BO.CO.

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DEFINITION.-The Orthogonal Projection of one straight line on another is the part of the latter intercepted between perpendiculars let fall on it from the extremities of the former,

B

Thus, if, from A and B, AL, BM be drawn perpendicular to CD, LM is the projection of AB on CD. It is clear

that the length LM depends only on the length of AB and the angle which AB makes with CD.

M

D

In Ex. 1, CO is the projection of AO on BO, and DO is

the projection of BO on AO. The theorem may, therefore, be stated thus:

If there be two straight lines, the rectangle contained by the first and the projection on it of the second is equal to the rectangle contained by the second and the projection on it of the first.

(4 Standard Theorem.)

2. In a trapezium the sum of the squares on the diagonals is equal to the sum of the squares on the oblique sides, together with twice the rectangle contained by the parallel sides.

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3. H is the point of intersection of the altitudes AP, BQ, CR of the acute-angled A ABC; show that

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Examine the case in which the triangle is obtuse-angled.

4. Prove, by means of a geometrical construction, that in obtuse-angled triangles the square on the side opposite the obtuse angle is equal to the sum of the squares on the other two sides, together with twice the rectangle contained by either of these sides, and the projection on it of the other side (Euc. II. 12).

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In figure 1, construct squares as in Euclid I. 47, and draw BRP, CSQI AC, AB.

As in Euc. I. 47, show that

BL=FS=AF+AQ=AB2+BAAS; and similarly, that

CL=CA2+CA.AR. Join, BH, CG, and use Euc. I. 4 and I. 39 to prove AQ=AP. Otherwise

In figure 2, on BC describe the square BCDE, and on BE, ED, CD describe the As BFE, EGD, CKD congruent with ABAC. Complete the square ABFH, and join HG, HK. Show that the square BCDE=the figure BACKDGEF, and show that AF=AB, DH=AC?, and that each of the parallelograms AK and EH is equal to the rectangle BA:AN.

Compare $ 20, Ex. 8.

5. Prove, by means of a geometrical construction, that in any triangle the square on the side opposite an acute angle is equal to the sum of the squares on the other two sides, dimi. nished by twice the rectangle contained by either of these sides, and the projection on it of the other side (Euc. II. 13).

[blocks in formation]

The constructions are similar to those of the previous Ex.

In figure 1, show that BL=FS=AB2- AB.AS, etc. Otherwise

In figure 2, AHFBC+GHKCD=GHFBCD+AHKC. Hence, show that AB2+AC2+2 AABC=BC2+EFHG+

AHKC+2AABC. Show, also, that EFHG=AHKC=AB.AN, and deduce the theorem.

6. If a rectangle and a square be equal in area, the square shall have the lesser perimeter. Use Euc. II. 14 to show that

H ABCD=BKHG, and show that

A
F

E
perimeter of AC=4FG,
perimeter of BH=4BG, etc.

Compare $ 33, Ex. 8.

B

D

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