Hello! I want help with a problem that I have tried to solve, but hopelessly :/ Here it is: http://s14.postimg.org/mazg3lcb5/geometry_unsolvable_problem.png AB = 6 cm AC = 9 cm BC = 10 cm BD = 4.2 cm Line segment AE is perpendicular to Line segment BF Required: Length of Line segment FC
Did you use the law of cosines to find the angles of the big triangle?
You can solve this using the law of cosines a few times. First find angle C in the big triangle use angle C to find the leg AD then find angle ADB (law of cosines on the triangle on the right side) that gives you enough info to find angle CBF and then the 3rd angle in the triangle CFB finally use law of sines there may be a faster way, but any way is good when it gets this complicated.
AEF and DEB are both right angle triangles. I would look at trying to closely compare them using complimentary angles.
Thanks for your help, I'll solve it as you said. We haven't studied the Sine rule and Cosine rule at school this year though. By the way this problem is on the lesson of interior and exterior angle bisectors, I don't find it related though.
I think there is a connection, I just can't remember it exactly. There is a name for the type of lines that intersect at right angles from the vertices of triangles
Off hand I don't know how to show that any of the angles are being bisected. But if we did know that line AD bisected angle A, then we can show the right triangles ABF and ABE are congruent. That would make AF=6 and FC=3
Well, for Line segment AD to bisect <(BAC), AB / AC should be equal to BD / CD because the internal bisector of an angle in the triangle divides the opposite side to the angle into two parts proportional to the other sides of the triangle, which isn't the case.
If BC was equal to 10.5 cm, that would be the case...
May be it should be 10.5 cm but it's just a typo :) Otherwise it should be solved using the Cosine rule and Sine rule as you stated earlier
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