Page images
PDF
EPUB

4. ABC is an isosceles triangle, having AB = AC. AD drawn to the base BC does not bisect A; prove that D is at unequal distances from B and C.

5. Prove the proposition with the same construction as in the text, but let DEG fall on the other side of DE.

[blocks in formation]

If two triangles have two sides of the one respectively equal to two sides of the other, but their bases unequal, the angle contained by the two sides of the triangle which has the greater base shall be greater than the angle contained by the two sides of the other.

[blocks in formation]

Let ABC, DEF be two triangles, having AB = DE,

AC = DF, but base BC greater than base EF:

it is required to prove L A greater than L D.

If ∠ A be not greater than 4 D, it must be either equal

D.

A is not = ∠D, for then base BC would be

to LD, or less than

But

=

base EF,

I. 4

which it is not.

Нур.

And ∠ A is not less than 4 D, for then base BC

would be less than base EF,

I. 24

which it is not.

Нур.

... A must be greater than 4 D.

1. In the figure to the first deduction on I. 24, if AB is greater than BC, prove that ∠AOB is greater than 2 BOC.

2. ABCD is a quadrilateral, having AB = CD, but the diagonal BD greater than the diagonal AC; prove that DCB is greater than 2 ABC.

3. ABCD is a quadrilateral, having AB = CD, but 2 BCD greater than ABC; prove that 2 DAB is greater than 2 ADC. 4. ABCD is a quadrilateral, having AB = CD, but - DAB greater than 2 ADC; prove that 2 BCD is greater than 2 ABC.

Dis the middle
ADB is acute.
Dis any point

5. ABC is a triangle, having AB less than AC. point of BC, and AD is joined; prove that 6. ABC is an isosceles triangle, having AB = AC. such that BD is greater than DC; prove that AD does not bisect A.

7. ABC is a triangle, having AB less than AC, and AD is the median drawn from A; prove that G, any point in AD, is nearer to B than to C.

PROPOSITION 26. THEOREM.

If two angles and a side in one triangle be respectively equal to two angles and the corresponding side in another triangle, the two triangles shall be equal in every respect ; that is,

(1) The remaining sides of the one triangle shall be equal to

the remaining sides of the other.

(2) The third angles shall be equal.

(3) The areas of the two triangles shall be equal.

[blocks in formation]

In As ABC, DEF let L ABC = L DEF, L ACB

= L DFE, and BC = EF:

it is required to prove AB = DE, AC = DF, L A = L D,

[blocks in formation]

If AB be not = DE, one of them must be the greater. Let AB be the greater, and make BG = DE; and join GC.

[blocks in formation]

...L GCB = ∠ACB, which is impossible.
Hence AB is not unequal to DE, that is, AB = DE.

I. 3

Const.

Нур.

Нур.

I. 4.

Нур.

[blocks in formation]

.:. AC = DF, L A = L D, ∆ ABC =

[blocks in formation]
[blocks in formation]

In As ABC, DEF let L B = L E, ∠ C = LF, and

[blocks in formation]

it is required to prove BC = EF, AC = DF, L BAC = L EDF,

ABC = DEF.

If BC be not = EF, one of them must be the greater. Let BC be the greater, and make BH = EF; and join AH.

I. 3

E

[blocks in formation]

... LAHB = L ACB, which is impossible.
Hence BC is not unequal to EF, that is, BC = EF.

Now in As ABC, DEF,

{

AB = DE

BC = EF

LB = LE;

Нур. Const.

Нур.
I. 4

Нур.

I. 16

Нур.

Proved

Нур.

.. AC = DF, L BAC = LEDF, ABC = A DEF. I. 4

1. Prove the first case of the proposition by superposition. 2. The straight line that bisects the vertical angle of an isosceles triangle bisects the base, and is perpendicular to the base.

3. The straight line drawn from the vertical angle of an isosceles triangle perpendicular to the base, bisects the base and the vertical angle.

4. Any point in the bisector of an angle is equidistant from the arms of the angle.

5. In a given straight line, find a point such that the perpendiculars drawn from it to two other straight lines may be equal.

6. Through a given point, draw a straight line which shall be equidistant from two other given points.

7. Through a given point, draw a straight line which shall form with two given intersecting straight lines an isosceles triangle.

PROPOSITION A. THEOREM.

If two sides of one triangle be respectively equal to two sides of another triangle, and if the angles opposite to one pair of equal sides be equal, the angles opposite the other pair of equal sides shall either be equal or supplementary.

In As ABC, DEF let AB = DE, AC = DF, L B = LE:

it is required to prove either LC = ∠ F, or∠C+∠F = 2 rt. Ls.

[blocks in formation]

··· ∆s ABC, DEF are equal in all respects, and LC = LF.

[blocks in formation]

Нур.
Нур.

I. 26

E

F

G

I. 23

At D make L EDG = L BAC;

« PreviousContinue »