Just as an introduction for SDP-partners, this is a

• theoretical ppt on properties of triangles

• in which

• first, 3 properties are formulated and visualised

(recalling or introducing new concepts)

• afterwards, 2 of these properties are proved

• while building up the proves interactively,

pupils draw and write on prefab-sheets which

combine multiple slides into 1 page (sheets are

included here but not translated)

7.3

Properties

of triangles

A

C B

A middle parallel of a triangle

(a line connecting the middles of two sides of

the triangle)

is // with the third side and has half of its length

1)

A

C B

Two medians of a triangle

(lines through angle and middle of opposite side)

divide each other in 2 parts which are in the ratio of

2 to 1

2)

2

1

In a rectangular triangle the height onto the hypothenuse

is middleproportional

between the line segments

3)

in which it divides the hypothenuse

A

C B

h

x

y

h2 = x . y

Een middenparallel van een driehoek (een lijnstuk dat de ………………………………………………………………

…………………………..) is // met de derde zijde en …………………………….

Gegeven: ABC met M het midden van [AB] en N het midden van [BC]

Te bewijzen: MN // AC en …………………...

Bewijs:

Beschouw ABC en MBN :

• B = …………………..

• = ……… (……………..) ABC ……………………

M = A

AB wordt door MN en AC gesneden

volgens ……………………………….. ………………………………

…………………… |MN| = ………………

A

C B

........BM

BA

.................... MN

AC

A middle parallel of a triangle (a line which

…………………………………………………………

is // with the third side

……………… Given: ABC with M the middle of [AB]

and N the middle of [BC]

To be proved: MN // AC and …………………...

A

C B

connects the middles of two sides of the triangle)

and has half of its length

M

N

|MN| = |AC| 21

1) 2)

A

CB

M

N

Prove:

Consider ABC and MBN :

• = …………………..

•

B in common

.............. BM

BA2

BN

BC (……….)

ABC …………. MBN

)(sideside

anglesideside

1)

A

CB

M

N

ABC …………. MBN

1)

...ˆ M A

AB is cut by MN and AC according to ……………………………….. equal corresponding angles

MN // AC

A

CB

M

N

ABC …………. MBN

2)

..............MN

AC......

BM

BA= 2

....MN

AC= 2

|MN| = |AC| 21

In een rechthoekige driehoek is de hoogte op de schuine zijde ………….…………………………………. tussen de lijnstukken waarin ze de schuine zijde verdeelt. (zie p A.18)

Gegeven: rechthoekige ABC met BH de hoogetlijn op [AC]

Te bewijzen: |BH|2 = |AH|.|HC|

Bewijs:

Beschouw AHB en BHC :

• A = …………………………………...

• C = …………………………………... (……….) AHB ……………………

A

C B

..............AH

BH

………………………..

of

………………………..

In a rectangular triangle the heigth onto the hypothenuse is

………….………… between the line segments in which it

divides the hypothenuse

middleproportional

A

C B

Given: rectangular ABC with BH the perpendicular onto [AC]

To be proven: ……………..H

|BH|2 = |AH|.|HC|

A

C B

HProve:

Consider AHB and BHC :

Name the angles in B: 21ˆˆ BenB 1

2.....................ˆ A1

ˆ90180 B

.....ˆ90 1 B2B

...........ˆ1 H 2

ˆ90 H (angle angle) AHB BHC

.....AH

BH

BH

HC

|BH|2 = |AH|.|HC|

… and now exercises …