1951 AHSME Problems/Problem 27

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Problem

Through a point inside a triangle, three lines are drawn from the vertices to the opposite sides forming six triangular sections. Then:

$\textbf{(A)}\ \text{the triangles are similar in opposite pairs}\qquad\textbf{(B)}\ \text{the triangles are congruent in opposite pairs}$ $\textbf{(C)}\ \text{the triangles are equal in area in opposite pairs}\qquad\textbf{(D)}\ \text{three similar quadrilaterals are formed}$ $\textbf{(E)}\ \text{none of the above relations are true}$

Solution

Say we draw 3 different types of triangles as the following

1.An equilateral triangle with side lengths $x$

2.An isosceles triangle with side lengths $x+3,x+3,x-1$

3.A scalene triangle with sides $x+7,x-3,x+4$

1. Gives us 6 congruent $30-60-90$ triangles, which means

$\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)}$ are all true. therefore 1. does not give any unique criteria.

$\textbf{(A)}$-Since the triangles are congruent with each other, they are similar by a ratio of $1$.

$\textbf{(B)}$-Since all the triangles are congruent they must always be congruent in any way.

$\textbf{(C)}$-The triangles are all congruent meaning that they always have equal area.

$\textbf{(D)}$-The top $2$ triangles, bottom-right $2$ triangles, and bottom-left $2$ triangles all make 3 identical congruent quadrilaterals.

2. Gives us 3 pairs of adjacent congruent triangles, which means

$\textbf{(E)}$ is the only choice that works.

$\textbf{(E)}$-All of the relationships don't work so we must choose the last choice, no relations.

3. Is not needed as we have found that all of the relationships given are not necessary all the time, so $\textbf(E)$ is the only choice that works.

So the answer is $\fbox{\textbf{(E)}}$

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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