I refuse to believe!

Test Preparation SAT Preparation
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<p>Here's me, doing practice tests trying to get an 800 in math. And yet again I come upon a problem which reveals a basic deficiency.</p>

<p>In a right triangle ABC, AB is perpendicular to AC and ABC = 60 degrees. Line BD bisecs ABC. What is AD in terms of BC?</p>

<p>I came up with two different methods, but they give different answers!</p>

<p>The first one is to think about triangle ABD, which is an inverse 30-60-90 because it has the 30 and 60 degree angles on opposite sides. AB = BC / 2. AB = AD * (3)^(1/2). AD = BC / (2*3^(1/2))</p>

<p>The second one is to find AC, and AD = AC / 2 because the angle is bisected. AD = AC / 2 = (BC / 2) * (3)^(1/2)</p>

<p>So AD = BC / (2*3^(1/2)) = (BC / 2) * (3)^(1/2)</p>

<p>I know it looks ugly on text, but try to write it pen and paper and you'll see what I mean. I've been fiddling around with this for half an hour and still can't find the source of the error.</p>

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<p>How did you get AB = AD * (3)^(1/2)? I don’t think I’m following what you’re saying correctly, but I tried it out and for some reason I got AD = ((3)^(1/2) / 4)*BC, which is probably wrong…</p>

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<p>I’m a little confused and probably wrong. Though, when you say ‘line bd’ then couldn’t point D be located anywhere on the line (which continues forever)? This would make for an infinite number of answers.</p>

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<p>^No, because I think he said that BD has to bisect ABC, which means that angle ABD has to be 30 degrees…</p>

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<p>True, but no matter where point D is located, the line will still create 2 30degree angles.</p>

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<p>^Oh okay, I get what you mean now. I made assumption that the point D has to be on the right triangle. In that case then, there will indeed be infinite answers.</p>

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<p>Cool…I figured i was probably wrong. So does the math section of the SAT test have trick questions?</p>

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<p>^xrCalico23</p>

<p>It’s a 30-60-90 triangle. If the side opposing the 30 degree angle is X then the side opposing the 60 degree angle is X * (3)^(1/2)</p>

<p>I formulated the second method wrong. Should be AD = AC / 2 = ((BC / 2) * (3)^(1/2))/2 = ((3)^(1/2)/4) * BC like what you got.</p>

<p>So AD = BC / (2*3^(1/2)) = ((3)^(1/2)/4) * BC, which is still wrong. </p>

<p>The book sides with the first method.</p>

<p>^cb0610</p>

<p>There’s an illustration that shows point D falls exactly in the middle of AC.</p>

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<p>Israelexwired, thanks for the clarification, I think I know why the answer I gave was wrong now. If you make angle B something really big, you can see that clearly, AD would not equal to AC/2 even though angles ABD and DBC are equal. Therefore, you have to rely on the other method.</p>

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<p>By drawing line D, you create two triangles that both share side AB. If we give AB an arbitrary length of K, then BC = 2K and AD = Ktan(30). tan(30) = 1/sqrt(3), so AD = K/sqrt(3). 2<em>AD = 2K/sqrt(3), and 2K = BC, so BC/sqrt(3) = 2</em>AD. BC/(2*sqrt(3)) = AD. Isn’t this what you got in the first post, as it would be right.</p>

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<p>What it comes down to is the faulty assumption that AD = AC / 2.</p>

<p>So a bisecting angle doesn’t bisect the opposing side? I’m amazed and astonished at this. I’ve gotten so many questions right having taken this for granted.</p>

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<p>I believe that is only the case in Isosceles triangles.</p>

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<p>To the OP and whoever else that’s interested: The answer is radical(3)/3.
Im 100% sure. I don’t really feel like explaining though.</p>

<p>And Yes, you can trust me.</p>

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<p>Fresh101 is right, AD is Root 3 / 3.</p>

<p>I’m probably not the best at explaining it, but here it goes:</p>

<p>1) Draw the triangle (BAC as the right angle in this case)</p>

<p>2) Let AB = 1 , AC = Root 3 , and BC = 2
(If you don’t understand this part then you need to go over the special triangles)</p>

<p>3) Because BD bisects angle ABC (which is given to be 60 degrees), we can say that < ABD is 30 degrees</p>

<p>4) Since side AB = 1, and < ABD is 30 degrees, length AD must be 1 / Root 3</p>

<p>5) Rationalize the denominator, and the length of AD in terms of the original triangle is Root 3 / 3</p>

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<p>^with all due respect to your explanation, there’s a much simpler way. However, I have no motivation whatsoever. <em>sighs</em></p>

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<p>I’ll try to explain this as concisely as possible.</p>

<p>B
|.
|…\
|…
|…
|…
|…
A___<strong><em>D</em></strong>C</p>

<p>BD bisects angle ABC (splits it into two 30 degree angles). We know that triangle ABC is a 30-60-90 triangle, with AB being the shorter side. Since the shorter side is half the length of the hypotenuse, AB = BC/2.</p>

<p>Triangle ABD is another 30-60-90, with AB as the longer side this time. Since long side = short side * sqrt(3), short side, or AD, = BC/(2 * sqrt(3)).</p>

<p>I only briefly glanced at the question, and I gotta get back to homework, so I may have misread something. Lemme know if I did, and I hop this helps.</p>

<p>got2surf</p>

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<p>So, how is SQRT(3) / 3 in terms of BC ?</p>

<p>I got the same thing as got2surf which simplifies to [BC * sqrt(3)] / 6</p>