Intercept Proofs (1 Viewer)

[Lukybear]

1. D,E and F are the mid-points of AB, BC, and CA respectively. Prove that
**a) FG=GD
**b)AG=GE

2.) PQRS is a trapezium with PQ // SR. If A, B and C are the mid-points of SP, PR and QR respectively prove that
a)AB//SR
b)BC//PQ
**c) the points A, B and C are collinear.

5) PQRS is a trapezium with PQ//SR. If A and B are the mid-points of SP and RQ respectively prove that
a) AB is parrallel to PQ and SR
**b) AB = 1/2(PQ+SR)

The asteriksked one i need help with. Thanks.

Also just with a general question; is it just universally known that:
the straight line joining the mid-points of two sides of a triangle is
a) parallel to the third side
B) one half of the length of the third side
or must i prove this?

 

[kurt.physics]

1. D,E and F are the mid-points of AB, BC, and CA respectively. Prove that
**a) FG=GD
**b)AG=GE


The asteriksked one i need help with. Thanks.

Also just with a general question; is it just universally known that:
the straight line joining the mid-points of two sides of a triangle is
a) parallel to the third side
B) one half of the length of the third side
or must i prove this?

Regarding your general question, yes, it is a well known theorem.

1(a)

(line joining the midpoint of two sides is parallel to the third)

In triangles AFG and ACE, both share angle CAE and because FD//BC both other corresponding angles are equal.



Furthermore

(corresponding sides in the same ratio)

Similarly in triangles AGD and AEB,



But it is given that



ie

(b)In the first part we proved ,

Using the fact that their sides are in the same ratio (which is 2):



But

 

[kurt.physics]

2.) PQRS is a trapezium with PQ // SR. If A, B and C are the mid-points of SP, PR and QR respectively prove that
a)AB//SR
b)BC//PQ
**c) the points A, B and C are collinear.

 

[kurt.physics]

5) PQRS is a trapezium with PQ//SR. If A and B are the mid-points of SP and RQ respectively prove that
a) AB is parrallel to PQ and SR
**b) AB = 1/2(PQ+SR)

In part (b) we will consider the areas in order to solve the problem

The general formula for the area of a trapezium is:



Now, at the point P drop a line such that it is perpendicular to the interval SR and name the point where it touches the line SR "C". Let the point of intersection of the line PC and AB be D.

Consider the two triangles PAC and PSD. As AB//SR, then the corresponding angles are equal and hence







We have just proved that the height of the trapezium PQBA is the same as ABRS. Let this height be "x"

So the area of PQBA is:



The area of ABRS is:



Now consider the large trapezium PQRS, its area is:



But the area of the larger trapezium is the sum of the two smaller trapeziums ie:

 

[Lukybear]

In part (b) we will consider the areas in order to solve the problem

The general formula for the area of a trapezium is:



Now, at the point P drop a line such that it is perpendicular to the interval SR and name the point where it touches the line SR "C". Let the point of intersection of the line PC and AB be D.

Consider the two triangles PAC and PSD. As AB//SR, then the corresponding angles are equal and hence







We have just proved that the height of the trapezium PQBA is the same as ABRS. Let this height be "x"

So the area of PQBA is:



The area of ABRS is:



Now consider the large trapezium PQRS, its area is:



But the area of the larger trapezium is the sum of the two smaller trapeziums ie:

Thanks so much. Your a genious!

 

[Lukybear]

Another General Question:

For the triangle in Question 4 in the attached file;

To prove ADF = DBE, are we suppose to let angle DAF be x and prove that this is equal to BDE?

Or are we suppose to prove similar for DFA to BCA. Then prove BED is simlar to BCA. Then say hence, BED ||| DFA

 

[kurt.physics]

Another General Question:

For the triangle in Question 4 in the attached file;

To prove ADF = DBE, are we suppose to let angle DAF be x and prove that this is equal to BDE?

Or are we suppose to prove similar for DFA to BCA. Then prove BED is simlar to BCA. Then say hence, BED ||| DFA

I am going to assume that DF and DE are perpendicular to AC and BC respectively (does the question mention this?)

In which case proving the two triangles is easy. The "test" you have to use is AAA, that is, all angles are equal (angle-angle-angle)

You could let , and then use the fact that the "big" triangle is right angled, and so the other angle is , and in both small triangles there is a right angle, and so the angle in each small one will be the opposite ie the triangle with will have its opposite angle and visa versa for the other triangle.

Because all the angles are equal they are similar.

But this question could be put slightly differently, DECF could be a square which would imply (which another property) that they are congruent.

Could you please post the full question so that i dont have to make too many assumptions =) I would then be able to fully clarify the specifics =)

 

[Lukybear]

And yes, you were right again, about them being right angles.

Thats what i thought, the way that you said. But heres the Q anywaz, from fitzpatrick 2u p113. It would be interesting to see how you set it out, but if your too busy, dw about it.

21 A ladder AB, 10m long, rests against a wall with B 6m from the wall.
a)How far up the wall does the ladder reach etc...
b)BD is 4m, prove that ADF ||| DBE
c) Calculate the distance DE and DF

 

[kurt.physics]

And yes, you were right again, about them being right angles.

Thats what i thought, the way that you said. But heres the Q anywaz, from fitzpatrick 2u p113. It would be interesting to see how you set it out, but if your too busy, dw about it.

21 A ladder AB, 10m long, rests against a wall with B 6m from the wall.
a)How far up the wall does the ladder reach etc...
b)BD is 4m, prove that ADF ||| DBE
c) Calculate the distance DE and DF

a) Pythagoras - 8 m

(b)There is the way you mentioned (ie name one angle whateva), but the more mathematically appealing "proof" is as follows.

















(c)There are a few ways again to do this, but the best way is to note that



and



To find DE:







To find DF: