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squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle.
Steps of the Demonstration to Case 1st.
(in which ad falls within the A.) 1. Prove that bc + BD = 2 BC X BD + Dc', 2. that Bc + (B0% + AD') = 2BC X BD +
(AD' + DC*), 3. that AB? + BC = 2 BC X BD + Ace*,
that ac alone < cb% + Bio by 2 BC X BD. * Obs. The learner should particularly mark the substitution of AB for BD + AD", and of ac for AD + DC.
Steps of the Demonstration to Case 2nd,
(in which ad falls without the A.) 1. Prove that ACB > right Z, 2. that :: AB? = AC + cb% + 2 BC X CD, 3. that AB' + BC = AC + (2 BC +2 BC X CD),
4. Prove that AB + BC = AC X (2 DB X BC)*,
i. e., that ac alone < AB% + Bc%, by 2 DB X BC. * Obs. Mark particularly the substitution of 2 DB X BC, for 2 BC + 2 BC X CD.
The last case, (in which the side ac of the A is I BC) is proved by xlvii. I.
PROPOSITION XIV. Problem. To describe a square which shall be equal to a given rectilineal figure.
1. Prove that BE X EF + EG? = GF%; and that this 2.
= Eh+ EG, 3. that be X EF, i. e., MBD = EH, 4. that rectilineal fig. A= Eh'.
( Argument ad absurdum).
Problem. To find the centre of a given circle.
Suppose that r is not in the centre of the o, but that some other point, as g, is the centre.
Then prove, on that supposition, 1. that in AS ADG, BDG) 2 ADG = _ BDG, 2. that :. they are both right Z S, 3. that _ FDB = L BDG; i. e. greater = less, which
shows the supposition to be false, 4. that r is the centre of the O.
(Argument ad absurdum).
Theorem. If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.
Steps of the Demonstration,
Suppose that AB falls without the O, and prove, on that supposition,
1. that _ DEB > Z DAE,
DF > DE; i. e. less > greater, which shows the supposition to be false, 4. that similarly AB does not fall upon the O; and
:: that it falls within it.
Theorem. If a straight line drawn through the centre of a circle bisect a straight line in it which does not pass through the centre, it shall cut it at right angles; and if it cuts it at right angles, it shall bisect it.
Part 1st is proved (from viii. 1.) by showing that (in AS AFE, BFE) Z AFE = _ BFE, and is that they are both right Zs.
Part 2nd is proved (from xxvi. 1.), by showing that in the same AS AF = FB.
Suppose that AC, BD do bisect each other in E.
(State that this is evidently impossible if one line pass through the centre, and the other not).
If neither of them pass through the centre, prove, on the above supposition,
1. that _FEA is a right Z, 2. that _ FEB is a right L, and :: = < FEA; i.e.
greater = less, which shows the above supposition to be false; and :: that AC, BD do not bisect each other.