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Teoirim 1 Tá uillinneacha rinnurchormhaireacha ar coimhéid
Select the proof required then click
mouse key to view proof.
Theorem 2 The measure of the three angles of a triangle sum to 1800 .
Theorem 3 An exterior angle of a triangle equals the sum of the two interior opposite
angles in measure.
Theorem 4
If two sides of a triangle are equal in measure, then the angles
opposite these sides are equal in measure.
Theorem 5
The opposite sides and opposite sides of a parallelogram
are respectively equal in measure.
Theorem 6
A diagonal bisects the area of a parallelogram
Teoirim 7
Tá tomhas na hullinne ag lár ciorcail cothrom le dhá oiread
thomhas na uillinne ag an imlíne ag seasamh ar an stual céanna
Theorem 8
A line through the centre of a circle perpendicular to a chord
bisects the chord.
Theorem 9
If two triangles are equiangular, the lengths of the corresponding
sides are in proportion.
Theorem 10
In a right-angled triangle, the square of the length of the side opposite to the right angle
is equal to the sum of the squares of the other two sides.
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Teoirim 1:
Tá uillinneacha rinnurchomhaireacha ar comhéid
T&T 2 ltch 205:
Dein cliceáil chun na céimeanna
A fheiscint diaidh ar ndiaidh
1
4
2
3
Le Cruthu 1 = 3 and
2 = 4
1 + 2 = 1800
Cruthú:
…………..
2 + 3 = 1800

…………..
Líne Díreach
Líne Díreach
1 + 2 = 2 + 3
1 = 3
Ar an modh céanna
2 = 4
Q.E.D.
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Teoirim 2:
The measure of the three angles of a triangle sum to 1800 .
Use mouse clicks to see proof
Given:
Triangle
To Prove:
1 + 2 + 3 = 1800
Construction: Draw line through 3 parallel to the base
4 3 5
3 + 4 + 5 = 1800
Proof:
Straight line
1 = 4 and 2 = 5 Alternate angles

1
2
3 + 1 + 2 = 1800
1 + 2 + 3 = 1800
Q.E.D.
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Theorem 3:
An exterior angle of a triangle equals the sum of the two interior
opposite angles in measure.
Use mouse clicks to see proof
3
4
To Prove:
Proof:
1
2
1 = 3 + 4
1 + 2 = 1800
…………..
2 + 3 + 4 = 1800
Straight line
…………..
Theorem 2.
1 + 2 = 2 + 3 + 4
1 = 3 + 4
Q.E.D.
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Theorem 4:
If two sides of a triangle are equal in measure, then the angles
opposite these sides are equal in measure.
a
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3 4
Given:
Triangle abc with |ab| = |ac|
To Prove:
1 = 2
Construction: Construct ad the bisector of bac
Proof:
b
d
In the triangle abd and the triangle adc
3 = 4
…………..
|ab| = |ac|
|ad| = |ad|
…………..
…………..
Construction
Given.
Common Side.
The triangle abd is congruent to the triangle adc

1 = 2
Constructions
2
1
………..
SAS = SAS.
Q.E.D.
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c
Theorem 5:
The opposite sides and opposite sides of a parallelogram
are respectively equal in measure.
Use mouse clicks to see proof
b
c
3
Given:
Parallelogram abcd
To Prove:
|ab| = |cd| and |ad| = |bc|
4
and
Construction:
1
a
2
d
Proof:
abc = adc
Draw the diagonal |ac|
In the triangle abc and the triangle adc
1 = 4 …….. Alternate angles
2 = 3 ……… Alternate angles
|ac| = |ac| …… Common
The triangle abc is congruent to the triangle adc

………
ASA = ASA.
|ab| = |cd| and |ad| = |bc|
and
abc = adc
Q.E.D
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Theorem 6:
A diagonal bisects the area of a parallelogram
b
a
c
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d
x
Given:
Parallelogram abcd
To Prove:
Area of the triangle abc = Area of the triangle adc
Construction:
Proof:
Draw perpendicular from b to ad
Area of triangle adc = ½ |ad| x |bx|
Area of triangle abc = ½ |bc| x |bx|
As |ad| = |bc| …… Theorem 5
Area of triangle adc = Area of triangle abc
The diagonal ac bisects the area of the parallelogram
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Q.E.D
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Teoirim 7:
(T&T 2 ltch 265)
.
Tá tomhas na hullinne ag lár ciorcail cothrom le dhá oiread
thomhas na uillinne ag an imlíne ag seasamh ar an stual céanna
Dein cliceáil ar do luch chun gach céim a fheiscint
Le Cruthú:
| boc | = 2 | bac |
Tógáil:
Ceangail a le o agus sínigh é amach go r
a
2 5
o
Cruthú: Sa triantán aob
3
| oa| = | ob | …… Is gathanna iad araon

| 2 | = | 3 | …… Triantán comhchosach
1 4
r
c
b
| 1 | = | 2 | + | 3 | …… Uillinn sheachtrach
 | 1 | = | 2 | + | 2 |
 | 1 | = 2| 2 |
Mar an gcéanna sa triantán eile | 4 | = 2| 5 |
 | boc | = 2 | bac |
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Teoirim 8:
Líne trí lár ciorcail atá ingearach le corda ,
(T&T 2 ltch 268)
déroinneann sé an corda sin.
Cliceáil ar do luch chun gach céim a fheiscint
Tugtha:
Ciorcal le lár o
agus líne L ingearach le ab.
Le cruthú :
Tógáil
a
o
| ar | = | rb |
r
Ceangail a le o agus b go o
90 o
Cruthú: Sa dhá thriantán aor agus orb
aro = orb
Tá

L
………….
b
90 o ( Dronuillin )
|ao| = |ob|
…………..
Gathanna. (T aobhagáin)
|or| = |or|
…………..
Slios comónta ( Slios)
an triantán aor iomchuí leis an triantán orb
|ar| = |rb|
………
Rhs = Rhs.
DTS = DTS
Q.E.D
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Teoirim 9:
Má tá dhá thriantán comhuilleach (comhchosúil) tá
(T&T 2 ltch 282)
na sleasa comhfhreagracha I gcomhréir (sa chóimheas céanna)
Dein cliceáil ar do luch
Tugtha:
Dhá thriantán le huillinneacha ar comhéid
|ab|
Le Cruthú:
|de|
=
|ac|
|df|
|bc|
=
|ef|
Ar ab marcáil ax ar comhfhad le de.
Tógáil:
Ar ac marcáil ay ar comhfhad le df
x
4
a
d
2
2
5
y e
1
1 = 4
Cruthú

3

f
[xy] comhthreomhar le [bc]
|ab|
|ax|
|ab|
b
1
3
|de|
=
c
Constructions
=
|ac|
|ay|
Mar xy comhtreomhar le bc
|ac| Ar an modh
céanna =
|df|
|bc|
|ef|
Q.E.D.
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Teoirim 10:
I dtriantán dronuilleach tá fad an tsleasa
os comhair na dronuillinne cothrom le
suim fhad cearnaithe an dá shlios eile.
Úsáid do luch chun cliceáil
T&T 2 ltch 288
b
a
a
c
c
c
b
a
c
b
a
b
Constructions
Tugtha :
An Triantán abc
Le Cruthú:
a2 + b2 = c2
Tógáil
3 Thriantán dronuilleach mar léirithe
Cruthú:
Achar na  móire. = achar na  bige + 4(achar D)
(a + b)2
= c2
a2 + 2ab +b2
= c2 + 2ab
+ 4(½ab)
a2 + b2 = c2
Q.E.D.
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