The Angle Bisector and Equal Side Ratios
Date: 05/17/98 at 17:17:07 From: Jason Yerep Subject: Geometry Proof Please help me to solve this proof. Given: Triangle ABC Angle bisector BD B /|\ x / | \ y / | \ / | \ / | \ / | \ A/ | \C --------------- a D b Prove: x y --- = --- a b where x = side AB, a = segment AD, y = side BC, and b = segment DC.
Date: 05/19/98 at 10:06:06 From: Doctor Santu Subject: Re: Geometry Proof Dear Jason: This problem can be solved by comparing the areas of the two triangles. The crucial idea is that angle bisectors are equally distant from both sides of the angle. In other words, if you have an angle, and you draw the angle bisector, no matter which point on the bisector you pick, it will be the same distance from the two lines of the original angle. This is the main idea in the solution I came up with. Below is a figure drawn so that it doesn't look like an isosceles triangle. See the blue lines? They're perpendiculars from the point D to the two sides of the top angle. Since D is on the angle bisector, it's the same (perpendicular) distance from the two sides BA and BC. How do we know this? The two triangles with BD as a common side and containing the blue lines are congruent (SAA) so the blue lines, being corresponding parts, must be congruent. Now forget the blue lines for the moment -- imagine that they're not there. Ignore the red line, too, and what you have left is the original triangle you drew, with the line down the middle. Let's call the areas of the two triangles A1 and A2. Then: A1 = (1/2)*base*height = (1/2)*a*h where h is the altitude of the left triangle, the length of the red line. Also: A2 = (1/2)*b*h since the triangles have the same altitude, from the way their bases are collinear, and have a vertex in common. But you can figure the two areas in different ways, too. The triangle on the left can be turned around so that the blue line on the left is an altitude. Now its area is: A1 = (1/2)*x*k where k is the length of the blue line. In the same way, the area of the triangle on the right is: A2 = (1/2)*y*k So a*h = x*k and b*h = y*k. A simple calculation shows that: x/y = a/b Good luck, -Doctor Santu, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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