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Proof (of b).
angle AGH is equal to angle GHD (h).
angle AGH is equal to angle EGB.
(i) angle E G B is equal to angle G H D.
Q. E. D.
Proof (of c).
angle EGB is equal to angle GHD (i). add the angle BGH to each of these. Then, by Axiom 2a,
(k) angles EG B, BG H are equal to angles BGH, GHD.
But, by Euc. I. 13,
EGB, BGH are together equal to two right
... angles BGH, GHD are together equal to two right angles.
Q. E. D.
Given. The isosceles triangle ABC,
and DE parallel to B C. Required. To prove that angle ADE is equal to angle AED.
THEOREM (Euclid I. 30).
Repeat. The enunciations of Euc. I. 27, I. 29, and Axiom I.
Straight lines which are parallel to the same straight line are parallel to one another.
(a) The line AB parallel to the line EF
(b) And the line CD parallel to the line EF.
Required. To prove that line AB is parallel to line CD.
(c) Let the straight line GHK fall on the lines AB, CD, and EF.
Q. E. D.
AB is parallel to EF (a), and
.. by Euc. I. 29,
(d) angle A G H is equal to angle G H F. Again, EF is parallel to CD (b), and GHK falls upon them (c).
by Euc. I. 29,
(e) angle G H F is equal to angle HK D. .. by conclusion (d) and (e), and Axiom 1, angle A G H is equal to angle HKD.
But angles A G H and HKD are alternate angles. .. by Euc. I. 27,
AB is parallel to CD.
Given.-CD and EF each Gparallel to AB.
GH parallel to CD, and KL parallel to EF. Required. To prove that GH is parallel to KL.
PROBLEM (Euclid I. 31).
Given. The line AB,
and the point C. Required. To draw a line through C, parallel A to AB.
In A B take any point D.
To draw a straight line through a given point parallel to a given straight line. Particular Enunciation.
(a) At the point in the straight line CD, (by Euc. I. 23), make the angle DCE equal to the angle CDB (Fig. 2).
Produce the straight line EC to any point F (Fig. 3). Then EF shall be parallel to A B.
The straight line CD meets the two lines E F and A B, and makes the angle ECD equal to the alternate angle CDB (a).
. by Euc. I. 27,
EF is parallel to A B. Wherefore a straight line EF has been drawn through C, parallel to A B.
EXERCISE.-Given.-The triangle ABC.
Q. E. D.
Required. To draw a line which shall bisect BC (see Euc. I. 10, page 32), and be parallel to AC.
THEOREM (Euclid I. 32, First Part).
Repeat. The enunciations of Euc. I. 29. General Enunciation.
If a side of a triangle be produced, the exterior angle is equal to the two interior and opposite angles.
(Compare this with Euc. I. 16, page 56, which proves that the exterior angle is greater than either of the interior opposite angles.)
Given. The triangle A B C, having the side A C produced to D.
Required. To prove that angle A CD is equal to the two interior opposite angles, CA B, AB C.
(a) Through the point C draw (by Euc. I. 31) CE parallel to AB.