# Help with Geometry Problems?

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• Jul 15th 2010, 12:25 AM
GGGGGGGGGGGGGGGGGG
Help with Geometry Problems?
I'm totally stumped on how to do these:

1. Let P and Q be the points on the sides AB and BC of a tringle ABC respectively such that BP = 3PA and QC = 2BQ. Let K be the midpoint of the segment PQ. Prove that the area of the triangle AKC is equal to $\displaystyle 11S/24$, where S is the area of the triangle ABC.

2. Let ABC be a triangle such that angle ACB = 135°. Prove that:

$\displaystyle AB^2=AC^2+BC^2+\sqrt{2}\timesAC\timesBC$

Thanks
• Jul 15th 2010, 02:43 AM
Wilmer
Quote:

Originally Posted by GGGGGGGGGGGGGGGGGG
1. Let P and Q be the points on the sides AB and BC of a tringle ABC respectively such that BP = 3PA and QC = 2BQ. Let K be the midpoint of the segment PQ. Prove that the area of the triangle AKC is equal to $\displaystyle 11S/24$, where S is the area of the triangle ABC.

TRY it using right triangle 3-4-5:
Code:

 C (2)  Q (1)  B    (3)      P(1)A
Place point K on QP as per the problem.
Area ABC = 6 (easy part!)
You'll get area AKC = 11/4
11(6) / 24 = 11/4
So that worked! I'm too lazy to try "any" triangle (Sleepy)
• Jul 15th 2010, 04:55 AM
GGGGGGGGGGGGGGGGGG
Quote:

Originally Posted by Wilmer
Area ABC = 6 (easy part!)
You'll get area AKC = 11/4
11(6) / 24 = 11/4

Sorry stupid question, but how do you get that AKC = 11/4?
• Jul 15th 2010, 08:13 AM
Wilmer
Quote:

Originally Posted by GGGGGGGGGGGGGGGGGG
how do you get that AKC = 11/4?

Draw 2 perpendiculars: KD (D on AB) and KE (E on BC).
You'll now have right triangles KEC and KDA, plus rectangle BDKE to work with.
If you can't handle that, sorry but you're not ready for "any" triangle.
• Jul 15th 2010, 02:16 PM
GGGGGGGGGGGGGGGGGG
Thanks very much, anyone have a solution for number 2?