triangle ABC and I is a point in triangle ABC such that angle IBA > angle ICA , angle IBC > angle ICB , BI cut AC at B' , CI cut AB at C' . PRove the : BB' < CC'

please help me i'm very thanks

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- Oct 17th 2005, 02:19 AMmathgeometry
triangle ABC and I is a point in triangle ABC such that angle IBA > angle ICA , angle IBC > angle ICB , BI cut AC at B' , CI cut AB at C' . PRove the : BB' < CC'

please help me i'm very thanks - Oct 18th 2005, 03:09 AMticbol
" triangle ABC and I is a point in triangle ABC such that angle IBA > angle ICA , angle IBC > angle ICB , BI cut AC at B' , CI cut AB at C' . PRove the : BB' < CC' "

So there is a triangle ABC.

1) A line segment from corner B is drawn touching side AC at B'.

2) A line segment from corner C is drawn touching side AB at C'.

3) The two line segments, BB' and CC', intersect at point I.

4) Angle IBA (and so, angle B'BA) is greater than angle ICA (and so, angle C'CA).

5) Angle IBC (and so, angle B'BC) is greater than angle ICB (and so, angle C'CB).

Prove line segment BB' is shorter than line segment CC'.

Okay.

Combining or from 4) and 5), angle ABC is greater than angle BCA.

Hence, side AC is greater than side AB.

Then, area of triangle ABC = area of triangle ABC. [ :-) ]

Area of any triangle is (base)*(altitude)/2.

If base is side AC, area of triangle ABC = (AC)*(altitude from AC to apex B)/2.

This altitude, say, h1, is proportional to line segment BB'.

If base is side AB, area of triangle ABC = (AB)*(altitude from AB to apex C)/2.

This altitude, say, h2, is proportional to line segment CC'.

h1 is shorter than h2.

Because, with area of triangle ABC being constant or not changing, and AC being longer than AB, then h1 must be shorter than h2.

Hence, by proportion, BB' is shorter than CC'. --------------****

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In triangle IBC, IB is shorter than IC, because angle ICB is less than angle IBC.

But I cannot connect that fact with the other fact that AB is shorter than AC. - Oct 18th 2005, 06:53 AMmathQuote:

Originally Posted by**ticbol**

- Oct 19th 2005, 02:37 AMticbol
Think it over again. Maybe you'd get it this time.

Draw the figure. Maybe that will help.

The h1 is a line from B to AC. h1 is perpendicular to AC. Depending on how you sketched the figure, you'd see that h1 is almost the same as BB'.

Same as with h2 and CC'. - Oct 30th 2005, 03:38 PMmath
but h1 is differen BB'

h2 is differen CC'

h1 < h2 => BB' < CC' is wrong - Oct 30th 2005, 10:05 PMticbol
Do you understand proportional?

Who said H1 = BB', h2 = CC' ?

If you cannot see h1 is proportional to BB', you have to study more. - Oct 30th 2005, 10:38 PMmath
you say , h1/BB' = h2/CC' why ????

h1 proportional BB' is a theorem ??? - Oct 30th 2005, 10:42 PMticbol
That really shows you don't understand proportional.

"you say , h1/BB' = h2/CC' "

Did I say that?

Read again all my postings in this thread. See if you can find where I said that. - Oct 31st 2005, 12:46 AMmath
i'm understand h1<h2 but i don't understand h1<h2 --> BB'<CC