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Join the points F and B by the right line FB.
We thus get two As BEF and CEA, in which the sides B E and CE are equal (being the equal parts into which the line B C has been divided); A E and EF are equal (EF having been cut off of the same length as A E), and the < BEF is equal to the < CEA, because they are vertically opposite <s, formed by the intersection of the lines B C and AF. So that the two As BEF and C E A have two sides and the included < of the one equal respectively to two sides and the included < of the other.
These two As, therefore, are also equal to each other in
every other respect. (Prop. IV.)
Among these respects is that the ZECA is equal to the _'EBF.
But the < EBF is less than the < CBD. Therefore, the < ECA, which is equal to the < EBF, is also less than the < CBD.
It has thus been shown that the exterior < CBD is greater than that one of the interior Zs which is opposite to the side produced.
It still remains to be shown that the < CAB is also less than the ZCBD.
The < CAB cannot be compared directly with the 2 CBD; but, by producing the line C B to any point G, we get an < ABG equal to the < CBD (being vertically opposite to it), with which the < CAB may be compared in the same way as that in which the < ACB was compared with the ZCBD.
For this purpose bisect the line A B in the point L. Join the points C and L by the right line CL. Produce the line CL to any length. From the produced part cut off a part, LH, equal to the line CL. Join the points H and B by the right line H B.
We thus get two AS, ALC and BLH, in which
the side AL is equal to the side BL (A B being bisected in the point L); the side C L is equal to the side L H L H having been cut off of the same length
as CL); and the ZS ALO and BLH are equal, being vertically opposite. That is to
say, two sides and the < be-D tween them in the A ALC are
equal respectively to two sides and the between them in the A BLH.
It follows, therefore (Prop. IV.), that the As ALC and BLH are also equal in every other respect.
Among these respects is that the < CAL is equal to the < LBH.
But the < LBH is less than the Z A B G (being a part of it).
Therefore the < CAL is also less than the _ ABG.
But the _ ABG is equal to the vertically opposite <CBD.
Therefore the _CAL, being less than the Z ABG, is also less than the Z CBD.
It has thus been shown that the Zs ACB and CAB are each less than the exterior _ CBD, formed by producing the side A B.
Let this proposition now be demonstrated with the following variations of the figure :
1. Produce the side B A to any point M, and prove that the < ACB is less than the < CAM. The following directions will suffice: 1. Bisect the line AC in the point P. 2. Join B and P by a straight line.
3. Produce the line BP and cut off a part, P Q, equal to PB. 4. Join the points Q and A by
a right line. Then show that the ACPB is equal in all respects to the A P Q A. And so, that the ZACB is less than the Z CAM.
Then producing C A to N, show that the ZCB A is less than the <BAN, and therefore less than the < CAM.
2. Produce the side BC to any point X, and show that the Zs CAB and C B A are each of them less than the ZACX.
The rule for proceeding is always the following :-When a side of the A has been produced, the_ that is to be compared with the exterior _ is that of the A which is opposite the side produced. For making the necessary construction, bisect that side of the A which is one of the sides containing the exterior Z. Join the bisecting point with the opposite vertex of the A. Produce this line through the point of bisection, and cut off a part equal to the part within the A. Lastly, connect the point so obtained, by a right line, with the vertex of the exterior Z.
Any two angles of a triangle are less than two right angles.
For the proof of this proposition we must be able
to produce a given straight line to any length,
and we must know,1. That if one side of a Abe produced, the exterior
Z is greater than either of the two interior and
opposite Zs. (Prop. XVI.) 2. That if one straight line meets another, the
adjacent Zs which it forms with it are to
gether equal to two right Zs. (Prop. XIII.) 3. That if the same quantity be added to each of
two unequal quantities, the sum of that quantity
and the larger of the two unequals is greater than the sum of that quantity and the smaller of the unequals. (Ax. IV.)
Let A B C be a A.
Take any two of its Z8, say CAB and CBA.
Produce one of the sides of the A, so as to form an exterior _ that
shall be adjacent to one or other of the two Zs CAB and CBA. As, for example, produce the side A B to the point D.
It follows from the last proposition that the ZCBD is greater than the < CAB. If the < CBA be added to each of these unequal <s, the sum of the <8 CBD and C B A will be greater than the sum of the Zs CAB and CBA.
But the Zs CBD and C B A are adjacent <s, formed by the straight line C B meeting the straight line A D, and are therefore equal to two right Zs.
Consequently, the Zs C AB and C B A are together less than two right Zs.
The proposition might have been proved equally well by either of the following constructions :
In each case we should have an exterior < greater than one or other of the two given <s of the A; enabling us to show that the sum of the two given <s, being less than that of one of them and the exterior Z, is less then two right Zs. Whichever two Łs of the A are selected, it does not matter which side of the A is produced, provided it is produced in a direction passing through the vertex of one of the two Zs fixed upon.
If one side of a triangle be greater than another, the angle opposite the greater side is greater than the angle opposite the less.*
For the construction requisite in this proposition,
we must be able, 1. From the greater of two straight lines to cut off
a part equal to the less. (Prop. III.) 2. To join two given points by a straight line.
(Post. I.) To prove the proposition by the aid of the con
struction, we must know that,1. If two sides of a A are equal, the Z8 opposite
to those sides are also equal. (Prop. V.) 2. If one side of a A be produced, the exterior <
is greater than either of the two interior opposite
Zs. (Prop. XVI.) Let ABO be a A in which the side C B is longer than the side C A.
We have to show that the Z CAB is larger than the <OBA.
From the greater side, CB, cut off a part, CD, equal to the less.
* The enunciation of this proposition, as it stands in the common editions of Euclid, is almost sure to mislead the beginner. It runs :-“The greater side of every Ais opposite the greater Z." This ambiguous statement is always to be interpreted in the form given above.