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Similarly KL is part, and equal to kl: and so on.

.. the į HKL = the hkl. [xl. 10.] In this way it may be shewn that the two polygons have their sides and angles severally equal, .. the polygons are equal in all respects.

24. Draw AX perp. to BC, and join ox.
Then ox is also perp. to BC. [Ex. 14, p. 418.]
Now, AX?. BC= (AO? + 0x2) BC? [1. 47.]

=AO2. BC? + OX?. BC?.

a* (0° + c) + bc? [Ex. 2, p. 336.] = a*b2 + 6co + cʻa?.

+

:: AABC = 4AX. BC= 1 Va’62 + b*c* + c*a?.

25. See the fig. on p. 426.

Let P and Q be opp. vertices of the octahedron, and A, B, C, D the remaining vertices, Then it may be easily proved that the fig. ABCD is a square,

.. AC? = ADP + DC? = 2AD?. Hence

AC = AD V2.

26. Let da, dB, dc be three conterminous edges of the cube, and D, a, b, c the vertices diametrically opposite to d, A, B, C respectively. Bisect Ab, bc, ca, aB, BC, CA in E, F, G, e, F, 9 respectively. Then the sides of the hexagon EfGeFg are clearly equal. And, if x be the middle pt. of cD, then EX, Xe are respectively par!. to and double of gc, CF; :. Ee is par! to and double of gF. Similarly Ee is par!. to and double of fg. Hence the pts. E, F, G, e, F, g are co-planar, and the hexagon EfGeFg is regular. [See Book iv. Prop. 15.]

[Four regular plane hexagons are obtained by bisecting all the edges, except those that meet (1) Aa, (2) Bb, (3) Cc, (4) Dd.]

27. Let o be the centre of the sphere.

Draw OC perp. to the plane of section; and take any point P on the line of section of the plane and sphere. Then

CP? = OP2 - OC (1. 47].
And since OP and oC are constant, CP is constant.

Hence all points on the line of section are equidistant from C. .. the section is a circle, of which C is the centre.

28. See fig. to p. 423.

Since the tetrahedron is regular, the perp' from the vertices meet the opp. faces at their centroids. [Ex. 11.]

Hence the perpendiculars meet at a point G, [p. 423.] where

Gg, = 4 Ag,

2 But 3Ag, = 2a? (Ex. 13). ..

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29. Let XY be the given plane, and AB the given st. line.

On AB as diameter describe a sphere. Then it follows from III. 31 that AB subtends a rt. angle at every point on the sphere.

Hence the required locus consists of the points common to the sphere and the plane, and is therefore a circle. [Ex. 27.]

30. Draw ON perp. to given plane, and in ON take A, so that rect. ON, OA=rect. OP, OQ=given constant. .. P, Q, A, N are concyclic. And LPNA is a rt. L. ..LAQO is a rt. L. .. locus of Q is a sphere described on OA as diameter.

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