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http://i118.photobucket.com/albums/o...s/triangle.jpg
In the above triangle ABC, line segment AD is 6, and I'm supposed to find the length of side BC.
I know I'm missing some basic knowledge about properties of triangles, or else this wouldn't completely baffle me. :confused:
Any help would be greatly appreciated. :)
Hello,Quote:
Originally Posted by thumpin_termis
I've modified your drawing a little bit so you can see how I did this problem:
(< this sign means angle):
<(CDB) = 2x° ==> <(ABD) = x°
Therefore triangle BAD is an isoscele triangle with the base AB. Thus AD = DB.
Triangle DCB is an isoscele triangle with the base DC. Thus DB = BC.
That means: 6 = AD = DB = BC
Hello, thumpin_termis!
Earboth is absolutely correct.
. . Here's the reason behind his solution.
Angle BDC = 2x is an exterior angle of triangle ABD.
. . Hence, it is the sum of the two non-adjacent angles.
That is: ./BDC .= ./BAD + /ABD
Hence: . . . 2x .= .x + /ABD
Then: . . /ABD .= .x
Therefore, triangle ABD is isosceles, BD = AD = 6, etc.
Awesome.
So lemme just make sure here: /ABD = x can be found from the original information given, since /BDC=./BAD + /ABD would be true in such situations. Did I get that right?
Hello,Quote:
Originally Posted by thumpin_termis
you got it.
Have a look at Soroban's post.
The exterior angle of a triangle is as great as the two non-adjacent angles of the triangle.
You can proof this "theorem" by using that the 3 angles in a triangle sum up to 180° and that a straight line forms an angle of 180° too.
Fantastic. Thank you very much guys.
angle 2x (D) = angle x (A) + angle () B .......... property of external angle to triangleQuote:
Originally Posted by CaptainBlack
hence angle B = x
hence side AD = Side BD
Similarly angle BDC = Angle BCD
hence side BD = Side BC = 6