8. Let XZ and YV be the two given planes intersecting in the st. line XY; and let these two planes be cut by the first of two par1. planes in AP, AQ and by the second in ap, aq.

shall the PAQ = the

paq.

Then

Because the par. planes PAQ, paq are cut by the plane XZ, .. AP and ap are par1. [xı. 16].

Similarly AQ, aq are par1.

.. the PAQ = the paq. [x1. 10.]

9. Let XY be the given plane, and AB the given st. line parl. to it. Let the plane ABba pass through AB and cut the given plane XY in the st. line ab: then shall ab be par1. to AB. For if not, AB and ab will meet if produced, since they are in the same plane ABba; but ab lies wholly in the plane XY; ..AB will meet the plane XY; which is impossible, for AB is given par1. to the plane.

Thus AB and ab, being in the same plane and not intersecting, are par1.

10. Let the two planes AY, CY pass one through each of the par1. lines AB, CD, and let XY be their common section. Then shall XY be par1. to AB and CD.

For if XY be not parl. to CD, these lines must intersect at Z, since they are in the same plane.

But XY is in the same plane ABYX; hence Z is in the plane ABYX and also in the plane ABDC;.. Z is in AB, their common section. That is, AB and CD intersect at Z; which is impossible, since they are par1.

Hence XY and CD not intersecting, and being in ‍the same plane, are par1.: .. XY is also par1. to AB [XI. 9].

11. Let ABYX, CDYX be two planes, having XY as their common section; and let PQ be a st. line parl. to both planes: then PQ shall be par1. to XY.

Through PQ take a plane, cutting the plane ABYX in ab, and the plane CDYX in cd; then ab and cd are each par1. to PQ, and therefore par to one another [Ex. 9, p. 418].

Hence, by Ex. 10, XY is par1. to ab and cd, and therefore to PQ [xI. 9].

12. Let AB, CD be the two st. lines, and P the given point. Take the planes containing AB and P, and CD and P; and let XY be their common section. Then XY shall be the line required.

For since P is a point in each plane, .. P lies in XY. And since XY is in a plane with AB, and also in a plane with CD, it intersects both of these lines.

13. Let X, Y, Z be the middle points of AB, BC, CD. Then AC is par1. to XY, a line drawn in the plane XYZ; .. AC is par1. to the plane XYZ; for if AC meet the plane XYZ at some point P, then P would be both in the plane AXYC and in the plane XYZ; that is, P would be in the common section XY, which is impossible, since AC and XY are par1.

Similarly BD is par1. to the plane XYZ.

Then EF is perp. to

14. Through E draw EF par1. to AB: the plane XY [XI. 8]; hence FEC is a right angle. But AEC is also a rt. angle: .. CE is perp. to the plane of EF, EA [x1. 4]. Now EF, EA, AB, EB are in the same plane [XI. 7]; .. CE is perp. to EB.

15. Let XYE, XYF be the two planes, having XY as their common section; and let BP, BQ be the common sections of these two planes with the plane of AP, AQ.

Then since AP is perp. to the plane XĒ, .. the plane of AP, AQ is also perp. to the plane XE [xi. 18].

Similarly the plane of AP, AQ is perp. to the plane XF.

Hence the plane of AP, AQ being perp. to the planes XE, XF, is perp. to XY their common section [xI. 19].

16.

Let XYE, XYF be the two planes, having the common section XY: and let A be a point in the plane XYE.

Then since AQ is perp. to the plane XF, .. the plane APQ is perp. to the plane XF. [xI. 18.]

And since AP is perp. to the plane XE, .. the plane APQ is perp. to the plane XĒ.

.. the plane APQ, being perp. to the planes XE, XF, is also perp. to XY their common section.

.. XY is perp. to PQ, a st. line which meets it in the plane

APQ.

17. Join AC, BD.

Then the six angles of the two As ABC, ADC, namely the <$ ABC, ADC, BAC, DAC, BCA, DCA are together equal to four rt. angles. [1.33.]

But the two BAC, DAC at the solid angle A are greater than the third BAD. [XI. 20.]

8

Hence, by addition, twice the sum of the AOX, BOX, COX is greater than the sum of the ▲ AOB, BOC, COA.

COX.

(ii) Let Oy be the common section of the planes AOB,

Then COB + BOY greater than ▲ COY [XI. 20]; to each add YOA.

Then COB + BOA greater than ▲ COY + YOA.

But ▲ YOA + YOX greater than ▲ AOX [XI. 20]; to each add L COX.

Then

COY + YOA greater than ▲ COX + AOX.

A fortiori COB + BOA greater than ▲ COX + L AOX. (iii) It has been proved that

▲ AOX + ▲ COX less than ▲ AOB + BOC;

similarly < BOX + LAOX less than BOC+L COA;

and

▲ COX + BOX less than ▲ COA+LAOB.

8

Hence, by addition, the sum of the AOX, BOX, COX is less than the sum of the AOB, BOC, COA.

8

In the plane COX

and on the side remote from C make the C'OX equal to the COX; and in OC, OC' take c, c' so that Oc Oc': then cc' will be bisected perpendicularly by OX at x. Through x in the plane AOB draw axb perp. to Ox meeting OA, OB in a, b. Join ac, bc'.

Then from the ▲ cxa, c'xb, we have ac = bc. [1. 4.]
Hence from the As aOc, boc', we have aOc = boc'. [1. 8.]
Now cOc' is less than the sum of boc, bọc;

S

That is, twice COX is less than the sum of $ COA, COB.

20. Let ABC be the Art. angled at C, O the middle point of AB, and P a point not in the plane of the A, such that

Then PO shall be perp. to the plane of ABC.

Join OC.

Then since ACB is a rt. angle, OA = OB = OC [III. 31].
Hence from the identically equal

8

< POA = POB =

▲ POA, POB, POC,
POC. [1. 8.]

But POA, POB, being adjacent are rt. angles; .. POC is also a rt. angle;

21.

and in the same plane,

.. PO is perp. to the plane ABC. [XI. 4.]

Let AB be a st. line drawn from the point A in the plane XY.

Draw BC perp. to the plane, and join AC. Then AC is the projection of AB on the plane.

Let AD be any other line drawn from A in the plane XY.
Then BAC shall be less than ▲ BAD.

Make AD equal to AC, and join BD, DC.

Then from the rt.-angled ▲ BCD, BD is greater than BC. And in the As BAC, BAD, we have BA, AC equal to BA, AD respectively, but base BC less than base BD;

.. ▲ BAC less than BAD. [1. 25.]

22. Let A, B be the points, and XY the plane.

Draw AF perp. to the plane, and produce it to E making FE equal to AF. Join EB cutting the plane in P. Join AP. Then AP + PB shall be a minimum.

For take any point R in the plane XY, and join AR, RB. If R is in FP (or FP produced) then AP + PB is less than AR+ RB. [Ex. 3, p. 243.] If not draw RQ perp. to FP and join AQ, QB.

Then it may be shewn AP, PB and AQ, QB lie in a plane perp. to XY, and that RQ is perp. to the plane AQB.

Hence AR is greater than AQ, and RB greater than QB. So that AP + PB is less than AQ + QB [Ex. 3, p. 243]; and AQ + QB less than AR + RB.

23. Let XYE and XYF be two planes having XY as their common section; and let PA, PB be drawn from a point P in the plane XYE so as to be equally inclined to the plane XYF.

From P draw PQ perp. to the plane XYF, and join AQ, BQ.

Then the PAQ = the PBQ. [Def. 4, p. 385.]

Hence the ▲ PAQ, PBQ are identically equal [1. 26]; .'. AP = BP ;

24.

.. the PAB = the PBA. [1. 5.]

Since PA is perp. to PB, PC, .. PA is perp. to the plane BPC [XI. 4]; and PX is drawn perp. to BC in that plane; hence it may be proved that AX is perp. to BC. [Ex. 3, p. 407.]

Similarly BY and CZ are respectively perp. to CA, AB.
.. XYZ is the pedal ▲ of the ▲ ABC.

25. Produce AO, BO, CO to meet BC, CA, AB respectively at X, Y,

Z.

Then because AP is perp. to PB, PC, .. AP is perp. to the plane PBC.

Hence the plane APXO, which passes through AP, is perp. to the plane PBC. [XI. 18.]

Similarly the plane APXO, which also passes through PO, is perp. to the plane ABC;

.. BC, the common section of the planes PBC, ABC, is the plane APXO [XI. 19];

.. AX is perp. to BC.

Similarly BY, CZ are respectively perp. to CA, AB;
.. O is the orthocentre of the ABC.

perp. to