Download kelompok 1 - WordPress.com

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Dessin d'enfant wikipedia , lookup

Golden ratio wikipedia , lookup

Steinitz's theorem wikipedia , lookup

Euler angles wikipedia , lookup

Duality (projective geometry) wikipedia , lookup

Simplex wikipedia , lookup

Noether's theorem wikipedia , lookup

Brouwer fixed-point theorem wikipedia , lookup

Line (geometry) wikipedia , lookup

Reuleaux triangle wikipedia , lookup

History of trigonometry wikipedia , lookup

Trigonometric functions wikipedia , lookup

Four color theorem wikipedia , lookup

Rational trigonometry wikipedia , lookup

Euclidean geometry wikipedia , lookup

Incircle and excircles of a triangle wikipedia , lookup

Integer triangle wikipedia , lookup

Pythagorean theorem wikipedia , lookup

Transcript
ANGGOTA :
1.
2.
3.
4.
5.
Maharani Asmara
(4101414004)
Ika Deavy M
(4101414013)
Shiyanatussuhailah
(4101414015)
Arum Diyastanti
(4101414017)
Novia Wulan Dary
(4101414019)
Theorem 7-5
The perpendicular bisectors of the sides of a
triangle intersect in a point O that is
equidistant from the three vertices of the
triangle.
Proof
Given: 𝜟 ABC with perpendicular bisectors l, l’,
and l’’,
Proof: l, l’, and l’’ are concurrent in a point O
and that OA = OB = OC.
C
L’’
l’
A
O
l
B
Statement
Reasons
L is the perpendicular bisector of
Given
L’ is the perpendicular bisector of
Given
L dan L’ intersect in a point O
If
OA = OB
A point a perpendicular bisector is
equidistant from the endpoints
OB = OC
Why?
OA = OC
Transitive property of equality
O is on the perpendicular bisector of
A point equidistant from two points is on the
perpendicular bisector of the segment
determined by those points.
O lies on L, L’, L’’ and OA = OB = OC
Statements 4 - 8
∦
then
L∦L’
Theorem 7-6
The angle bisectors of the angles of a
triangle are concurrent in a point I that is
equidistant from the three sides of the
triangle.
PROOF
Given : βˆ†π΄π΅πΆwith angle bisectors 𝑙, 𝑙 β€² π‘Žπ‘›π‘‘ 𝑙 β€²β€² .
Prove : 𝑙, 𝑙 β€² π‘Žπ‘›π‘‘ 𝑙 β€²β€² are concurrent in a point I that
is equidistant from the three sides of the triangle.
𝑙
𝑙′
𝑙 β€²β€²
If you were to construct a triangle and its three
altitudes, you would see that the lines containing
the altitudes are concurrent.
Theorem 7-7
The lines that contain the
altitudes of a triangle intersect
in a point.
Definition 7-1
A medians of a triangle is a segment joining a
vertex to the midpoint of the opposite side
Theorem 7-8
The medians of a triangle intersect in a
point that as two thirds of the way from
each vertex to the opposite side
Theorem 7 - 9
If the measures of two angles of a triangle are
unequal, then the length of the side opposite
the smaller angle is less than the length of the
side opposite the larger angle.
Proof
Given: 𝜟 ABC with m ∠ B < m ∠ A
Prove: AC < BC
C
D
A
B
Statement
Reasons
m∠B<m∠A
There exists a point D on
∠ BAD = m ∠ B
β‰…
Given
so that m
Protractor Postulate
If two angles of a triangle are congruent,
then the sides opposite them are congruent
AD = BD
Why?
AC < AD + DC
Why?
AD + DC = BD + DC
Addition of equals property
BD + DC = BC
Definition of between for points
AC < BC
Substitution Principle
Theorema 7-10
If the lengths of two sides of a triangles are
unequal then the measure of the angle
opposite the shorter side is less than the
measure of the angle opposite the longer side
PROOF
Coba kita buat segitiga sembarang, misalnya segitiga
ABC seperti gambar berikut ini.