Here's an interesting little math puzzle for you all. The first to solve it will get a prize.
disclaimer: just pay shipping and handling
GIVENS:
You are given an isosceles triangle.
CHALLENGE:
You must draw two congruent right triangles within the given isosceles triangle, and then explain why what you drew fits the requirements.
THE CATCH!?!?
You may use two and only two additional lines to form these two congruent right isosceles triangles. In other words, you may not draw a line from the top vertex to the midpoint of the base as a solution, because that solution uses only one line. Furthermore, the endpoints of the two lines you draw must fall on the lines of the original triangle.
disclaimer: just pay shipping and handling
GIVENS:
You are given an isosceles triangle.
CHALLENGE:
You must draw two congruent right triangles within the given isosceles triangle, and then explain why what you drew fits the requirements.
THE CATCH!?!?
You may use two and only two additional lines to form these two congruent right isosceles triangles. In other words, you may not draw a line from the top vertex to the midpoint of the base as a solution, because that solution uses only one line. Furthermore, the endpoints of the two lines you draw must fall on the lines of the original triangle.
You are weak. WEAK.
For clarity's sake lets call the original triangle triangle ABC, where A is the top vertex, B is the left base vertex, and C is the right base vertex.
The way to solve this problem is to draw medians from the two base angles to the opposite sides.
Triangles DBF and ECF are congruent, right triangles; problem solved.
But the true question here is why are these triangles congruent and how can you know they are right triangles.
Well to start off, part of it is because when the you draw these two medians in any triangle, they divide each other into proportionate segments; a 1:2 ratio. This means that line DF and EF are one-third of line DC and EB, respectively, while BF and CF are two-thirds of BE and CD, respectively. We know this because if you draw a line to connect the two midpoints, you get two proportionate triangles, DEF and the pre-existing CBF. We know these are proportionate because they are similar. We know they are similar because line DE is parallel to line BC; and therefore, alternate interior angle pairs CBE, DEB and BCD, EDC are congruent, and having two consecutive corresponding congruent angles is a method of proof of similarity of triangles. Since line DE is half the distance of line BC (it is because it is parallel to the base and halfway up the triangle), and the triangles are similar, that means that all the other sides are likewise proportional to each other in a 1:2 ratio. *mindblows*
The medians drawn from the bases to the opposite sides of triangles are congruent (this is a theorem but I'm not going to bother to explain it. Think about it though, it makes sense.) Furthermore, if two equal things are divides by the same divisor, then the resulting quotient is equal in both cases. This means that if the two medians were both divided into a one-third segment and two-thirds segment - which they are - then the one-third segments are congruent and the two-thirds segments are congruent. FB is congruent to FC and DF is congruent to EF. Finally, angle BFD and angle CFE are vertical angles, and vertical angles are congruent to each other. Triangle BFD and triangle CFE are congruent by SAS. (Also noteworthy, triangle FBC is an isosceles right triangle)
This just leaves me to show why these triangles are right triangles. Its easier to do with coordinate geometry so I'll just do it that way.
Say the coordinate of A are (6,0), B(0,-2), C(0,2). You can use the midpoint formula to find the midpoints of AB and AC. They are (-1,3) and (1,3). If you use these points to find the slopes of the two medians, you will find that they are 1 and -1. Notice anything!?!?!??!? These are negative reciprocals, and lines whose slopes are negative reciprocals are perpendicular. The grand finale: perpendicular lines form right angles. Therefore, angles BFD and CFE are right angles, and triangles BFD and CFE are congruent right triangles.
Thank you and goodnight.
For clarity's sake lets call the original triangle triangle ABC, where A is the top vertex, B is the left base vertex, and C is the right base vertex.
The way to solve this problem is to draw medians from the two base angles to the opposite sides.
Triangles DBF and ECF are congruent, right triangles; problem solved.
But the true question here is why are these triangles congruent and how can you know they are right triangles.
Well to start off, part of it is because when the you draw these two medians in any triangle, they divide each other into proportionate segments; a 1:2 ratio. This means that line DF and EF are one-third of line DC and EB, respectively, while BF and CF are two-thirds of BE and CD, respectively. We know this because if you draw a line to connect the two midpoints, you get two proportionate triangles, DEF and the pre-existing CBF. We know these are proportionate because they are similar. We know they are similar because line DE is parallel to line BC; and therefore, alternate interior angle pairs CBE, DEB and BCD, EDC are congruent, and having two consecutive corresponding congruent angles is a method of proof of similarity of triangles. Since line DE is half the distance of line BC (it is because it is parallel to the base and halfway up the triangle), and the triangles are similar, that means that all the other sides are likewise proportional to each other in a 1:2 ratio. *mindblows*
The medians drawn from the bases to the opposite sides of triangles are congruent (this is a theorem but I'm not going to bother to explain it. Think about it though, it makes sense.) Furthermore, if two equal things are divides by the same divisor, then the resulting quotient is equal in both cases. This means that if the two medians were both divided into a one-third segment and two-thirds segment - which they are - then the one-third segments are congruent and the two-thirds segments are congruent. FB is congruent to FC and DF is congruent to EF. Finally, angle BFD and angle CFE are vertical angles, and vertical angles are congruent to each other. Triangle BFD and triangle CFE are congruent by SAS. (Also noteworthy, triangle FBC is an isosceles right triangle)
This just leaves me to show why these triangles are right triangles. Its easier to do with coordinate geometry so I'll just do it that way.
Say the coordinate of A are (6,0), B(0,-2), C(0,2). You can use the midpoint formula to find the midpoints of AB and AC. They are (-1,3) and (1,3). If you use these points to find the slopes of the two medians, you will find that they are 1 and -1. Notice anything!?!?!??!? These are negative reciprocals, and lines whose slopes are negative reciprocals are perpendicular. The grand finale: perpendicular lines form right angles. Therefore, angles BFD and CFE are right angles, and triangles BFD and CFE are congruent right triangles.
Thank you and goodnight.
