Question

In triangle ABC, O is a point on BC. Show that a straight line can be drawn parallel to AO that divides triangle ABC into 2 equal halves.

Answer 1

Is this the complete question??? Pls reconfirm this is the exact wording...

Answer 2

yes..its complete..

Answer 3

ok understood,

Now tell me whether a proof is to be given or should I give you the logic ??

Answer 4

anyone is great....u can xplain the logic...and give me the proof if u want (or if i dont understand) :)

Answer 5

ok, give me some time to think and get an answer......

Answer 6

ok..:)

Answer 7

sorry friend, i tried but nothing worthwhile occurred to me right now. Will keep thinking on this and if i come up with a solution, will post it here....

Answer 8

can u tell me on which chapter is this based
- mensuration
similar triangles or ???

Answer 9

i cant say..this problem came in my exam...

Answer 10

given that the measure of the triangle (area or perimeter) is bounded; then any increase in one portion has an equal decrease in the other up to the point were both sides reach an equilibrium.

Answer 11

it feels like more of a 'mean value theorum' then an actual way of determing specific values

Answer 12

is that a kind of proof or an intuitive thought :) lol :P

Answer 13

:) intuitive thought, i gots no way of proofing it ....

If the area of a Triangle is A; and we divide it into 2 unequal sections by a straight line (SL); then A = aSL^(-) + aSL^(+), where aSL^(-) denotes to the area left, and aSL^(+) denotes the area to the right.

Answer 14

ya...

Answer 15

the 'Rolles' thrm of triangular equality :)

Answer 16

but u are not given that area,perimeter is bounded...

Answer 17

the given problem is weird...i dont think geometry will be of help here :P

Answer 18

how can you have a triangle what is unbounded? do we have to redefine all the words we use or is there a threshold point at which the language can simply mean what it says?

Answer 19

the drawing says more than what is stated in the problem. As the drawing shows the line drawn as a perpendicular to one side. In the statement it seems that the point can be at any location on the line BC.

Answer 20

a given triangle ABC, no matter what is perimeter and area; remains constant for the triangle once defined

Answer 21

oops..mistake sorry..i didnt see ur meaning...i thought u were fixing the value of area and perimeter for all triangles.... :)

Answer 22

:) well, any triangle will do; just once you pick one ... keep it :)

Answer 23

yah...amistre that problem is "physically" conceivable...but i need a proof and i dont get any idea for a week! :)

Answer 24

@radar:O is any point on BC and AO may not be perpendicular to BC.

Answer 25

Then I don't believe the the statement. I think it is false. I have used various points and drew many lines parallel and they all do not for two equal (area wise) triangles!

Answer 26

for is form

Answer 27

there will be one out of many lines which are parallel to AO that divides the triangle in 2 equal halves

Answer 28

Then prove it.

Answer 29

thats the problem...i cant find a way to prove it...(i think the problem is correct. The real story behind this problem is that it appeared in my exam in which the question papers are taken..So this problem is what the students recall....it may be wrong...but I dont think so..)...

Answer 30

"In triangle ABC, O is a point on BC. Show that a straight line can be drawn parallel to AO that divides triangle ABC into 2 equal halves. "

There is no stipulation for the location of O

Answer 31

yup..thats what the problem was..O may be any point on BC...

Answer 32

So as I understand that one of the infinite possible parallel lines will divide the triangle into two equal halves but not two triangles???

Answer 33

yes...not necessarily atriangle..may be a quadrilateral+triangle...

Answer 34

similar to what you've shown in the diagram. Now I am beginning to believe the statement, but do not have a clue to prove it. I am in the same shape you're in. Sorry but I can't help you.

Answer 35

thnx for ur time, radar:)

Answer 36

I can see that if I could slide a parallel to AO line across that figure, at some point the area above and below that line would be equal. But can not prove it mathematically.

Answer 37

what if we make a construction:Join O with the point of intersection of the line (parallel to AO and which divides into 2 halves) with AB in the figure. U break the quadrilateral into triangles then...any ideas now?

Answer 38

No, maybe if there was a stipulation that AO is perpendicular to BC, but gosh O could be placed up near B or down near C. It looks like what amistre was saying that a triangle can be divided in two equal halves by moving a line across the surface until it reaches the "sweet spot" It now seems obvious, but the proof is subliminal

Answer 39

its good we "feel" the problem...but the bad thing is i cant find a proof. If I had a proof, what would it have: congruency? similarity? ceva's theorem? trigonometry?.....all approaches seem to fail ...

Answer 40

I don't feel the two equal area shapes would be congruent. Unequal angles etc.

Answer 41

if we ignore the shape and simply view the 'area' that the shape defines; then it is intuitive that at some given division, the amount of area will be in 2 equal halves :)

Answer 42

try to 'disprove' it :) ... forgot the name they use for that method

Answer 43

@amistre..ur statement does not say anything about AO and a lines parallel to it......
do u mean "contradiction"?

Answer 44

yes, proof by contradiction .....

Answer 45

this seems completely obvious. but just because something is obvious doesn't mean you cannot prove it.

Answer 46

OK...see this
Suppose a line not parallel to AO divides ABC into 2 halves.
Then we can find a point O' on BC such that AO' is parallel to our line.
But we can rename O' as O since O can be any point on BC.
Thus proved.
I dont know if this is a rigid proof...please comment if u see any "loopholes" in this..:)

Answer 47

Supoose a no line parallel to A0 divides in in half

Answer 48

@amistre: See my post....

Answer 49

the area on one side of the line will always be able to be defined in the shape of a triangle; at what point does the parallel line create half the area of the whole ?

Answer 50

i got no idea if your proof is valid, trying to write a proof is not something im good at :)

Answer 51

there are similar triangle proofs right? any line parallel to A0 makes a similar triangle at some intersection and the area of that new triangle is smaller than the whole of ABC; at what point is it half?

Answer 52

by similar i mean the triangle defined by AOB or AOC

Answer 53

no they may not be similar...

Answer 54

any dividing line that is parallel to AO at some point creates a similar triangle

Answer 55

The area of the similar triangle at some point is a division of the area of the whole triangle; therefore the similar triangle at some given point will be half of the whole.

Answer 56

If you can determine how to find the area of triABC; then you can determine how to find the area of one of the similar triangles created by a parallel line and measure it to determine if its area is half of triABC

Answer 57

what is it; Hoteps formula for the area of a triangle by using the perimeter?

Answer 58

heron's formula .:) (u r not well with geometry..i see..but ur ideas r good! :) )

Answer 59

not to jump in so late, but if you call the distance between O and O' say "x" is it not possible to find the area of the bottom triangle plus the piece A O x whatever on the other side? this will be a function continuous in x and it will be less than 1/2 the area of the triangle if x = 0 and the entire are of the triangle of x = OB and therefor must at some point be half the area of the triangle.

Answer 60

:)

Answer 61

i will see if i can find a nice formula for that area, but it will certainly be continuous that is for sure.

Answer 62

@amistre,satellite73: Draw a parallel line to AO through C, and slide this line up towards B. Let f(t) be the area below this line over time. This function is continuous. f(0)=0, and when this line reaches B at time x, f(x)=area(ABC) .

Therefore, by the Intermediate-Value Theorem there exists some time t between 0 and x such that f(t)=(area(ABC))/2.

Well this proof assumes that u r given the line parallel to AO and need to proof that it divides ABC int 2 halves.Is this what the question says. Or does it say the other way round?

Answer 63

.... i got know idea if that is correct or not :) srry

Answer 64

how do we know that the direction from O to B is the proper direction? if we split the area by AO and head in the smaller direction, we will never find a point.

Answer 65

i think i see your counting for that; yes, from point C to B sweeps the Area

Answer 66

maybe start at the point O and create 2 lines that sweep, one towards C and the other towards B .....

Answer 67

this would account for any notion that the shape of the area being defined for a quadralateral as opposed to a triangle

Answer 68

Create the line AO so that it IS AC and sweep it out maybe?