# Inequalities in Triangles

• May 13th 2007, 06:15 AM
Ruler of Hell
Inequalities in Triangles
Firstly, you guys must be wondering how come all threads started by me are always in the 'Urgent Homework Help' section. Well, that's me.... I always get reminded of homework a li'l late.

Secondly, Inequalities in Triangles is a topic I find difficult to understand at all. So, I'd kindly request you people to please post all the steps. Please! I understand the theorems of inequalities, however, I fail to be able to apply in the sums.

1. In triangle ABC, if AD is a median, prove that AB + AC > 2AD

2. In triangle ABC, AB > AC. From the greater side AB, AD is cut off equal to AC and DC is joined. Prove that -
i) Angle ACD = 1/2 (Angle B + Angle C)
ii) Angle BCD = 1/2 (Angle C - Angle B)

Thanks a lot! Thanks!
• May 13th 2007, 07:46 AM
earboth
Quote:

Originally Posted by Ruler of Hell
...
Secondly, Inequalities in Triangles is a topic I find difficult to understand at all. ...
1. In triangle ABC, if AD is a median, prove that AB + AC > 2AD
...

Hello,

to #1:

1. Draw a sketch.
2. Inflect the triangle ABC over the point D (not sure if this is the correct expression; see attachment)

3. You'll get a parallelogram ABA'C. The diagonal of the parallelogram is 2*(AD)
4. Consider the triangle ABA' with BA' = AC

5. According to the triangle inequality
AB + BA' > AA' = 2*(AD)
• May 13th 2007, 07:59 AM
ThePerfectHacker
Quote:

Originally Posted by earboth
2. Inflect the triangle

Reflect, I think is the correct expression.

Anyway, you have a nice proof, nicer than the once I was thinking of. The nice think about your proof is that you do not use the fact that the line is a median. Thus it works for any line inside a triangle!

---
Quote:

Originally Posted by Ruler of Hell
2. In triangle ABC, AB > AC. From the greater side AB, AD is cut off equal to AC and DC is joined. Prove that -
i) Angle ACD = 1/2 (Angle B + Angle C)
ii) Angle BCD = 1/2 (Angle C - Angle B)

You draw a picture and follow what I do.

1)Draw triangle ABC.
2)Mark off point D on AB.
3)Connect C and D.
4)Note AC = AD thus, triangle ACD is isoseles.
5)<ADC is exterior angle for triangle DCB.
6)Thus, <ADC = <DCB + <DBC
7)Thus, <ACD + <ADC = (<ACD+<DCB)+<DBC
8)Thus, <ACD + <ADC = <C + <B
10)Thus, 2<ACD = <C + <B
Q.E.D.
• May 13th 2007, 08:12 AM
Ruler of Hell
@earboth: Amazing method, thanks a lot! Thanks a tonne!
@ThePerfectHacker: The method you've shown is so much easier than what I've found. Thanks a lot!

Thanks a lot guys, thanks a lot!:D :D :D Thank you!!
• May 13th 2007, 08:17 AM
earboth
Quote:

Originally Posted by Ruler of Hell
...
Secondly, Inequalities in Triangles is a topic I find difficult to understand at all. ...
2. In triangle ABC, AB > AC. From the greater side AB, AD is cut off equal to AC and DC is joined. Prove that -
i) Angle ACD = 1/2 (Angle B + Angle C)
ii) Angle BCD = 1/2 (Angle C - Angle B)
...

Hello,

to #2:

1. Draw a sketch (watch out: The angle γ is the complete angle at C)
2. Triangle ADC is an isosceles triangle with the base CD.
i):
3. The red angles at the base are equal and have the value:
<(ACD) = (180° - α)/2
α = 180° - ß - γ. Now plug in this term into the previous equation:

<(ACD) = (180° - (180° - ß - γ))/2 = (ß + γ)/2
ii):
4. <(BDC) = γ - (ACD) = γ - (ß + γ)/2 = γ - ß/2 - γ/2 = (γ - ß)/2