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- sarora
MGMAT Math Question Bank
Can someone tell me how to solve this question through similar triangles?
- RonPurewal
- Students
- Posts: 19744
- Joined: Tue Aug 14, 2007 8:23 am
Well, you don't really need similar triangles for this problem.
The key is to notice that nothing originally fixes the length of AB, the 'other' leg of the right triangle (note AD is fixed = 5). Because the Pythagorean theorem holds, being able to figure out AB is equivalent to figuring out the desired BD (either can be found from the other with the Pythagorean theorem, because we've got the other leg = 5). Therefore, let's consider three different cases in order to try to prove 'insufficient':
* AB is really long, like 20+
* AB = 5
* AB is really short, like 2
Try drawing each of these cases; doing so will be very instructive and enlightening as to the ambiguities in the problem.
Statement (1)
This doesn't help by itself. If ABD is a right triangle as drawn, then an altitude can ALWAYS be drawn to the hypotenuse of the triangle, so any of the three possibilities listed above could work. Therefore, this choice is insufficient.
Statement (2)
Again, doesn't help by itself. You can ALWAYS connect ANY vertex of ANY triangle to the midpoint of the opposite side (try drawing random triangles to see that this is true). Therefore, all three triangle possibilities suggested above also work, so this one is insufficient too.
Together
If both statements are true, then triangles ACB and ACD are congruent. (You can tell this by symmetry, although, if you want a formal proof, you can get one from Side-Angle-Side). This means that all the triangles in the diagram are 45-45-90 triangles, making AB = 5 and BD = 5 * root(2).
The key is to notice that nothing originally fixes the length of AB, the 'other' leg of the right triangle (note AD is fixed = 5). Because the Pythagorean theorem holds, being able to figure out AB is equivalent to figuring out the desired BD (either can be found from the other with the Pythagorean theorem, because we've got the other leg = 5). Therefore, let's consider three different cases in order to try to prove 'insufficient':
* AB is really long, like 20+
* AB = 5
* AB is really short, like 2
Try drawing each of these cases; doing so will be very instructive and enlightening as to the ambiguities in the problem.
Statement (1)
This doesn't help by itself. If ABD is a right triangle as drawn, then an altitude can ALWAYS be drawn to the hypotenuse of the triangle, so any of the three possibilities listed above could work. Therefore, this choice is insufficient.
Statement (2)
Again, doesn't help by itself. You can ALWAYS connect ANY vertex of ANY triangle to the midpoint of the opposite side (try drawing random triangles to see that this is true). Therefore, all three triangle possibilities suggested above also work, so this one is insufficient too.
Together
If both statements are true, then triangles ACB and ACD are congruent. (You can tell this by symmetry, although, if you want a formal proof, you can get one from Side-Angle-Side). This means that all the triangles in the diagram are 45-45-90 triangles, making AB = 5 and BD = 5 * root(2).
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