14. construction construct kite benf given sides $\overline{be}$ and $\overline{en}$ and diagonal $\overline{bn}$. how many different kites are possible?

Question

Accepted Answer

# Explanation:
## Step1: Recall Kite Properties
A kite has two distinct pairs of adjacent sides equal. Given sides $\overline{BE}$ and $\overline{EN}$, and diagonal $\overline{BN}$. Let's denote the fourth vertex as $F$. In a kite, one diagonal is the perpendicular bisector of the other. Here, $\overline{BN}$ is a diagonal. The other diagonal (connecting $E$ and $F$) must be perpendicular to $\overline{BN}$ and bisect it (or vice - versa, but in this case, since we have sides $\overline{BE}$ and $\overline{EN}$, we consider the symmetry).

## Step2: Analyze the Position of $F$
We know that $BE = BF$ and $EN=FN$ (since in kite $BENF$, adjacent sides are equal: $BE = BF$ and $EN = FN$, or $BE=FN$ and $EN = BF$, but let's check the first case). The diagonal $\overline{EF}$ is perpendicular to $\overline{BN}$. For the given segment $\overline{BN}$ and the points $B$ and $N$ with $BE$ and $EN$ as sides, the point $F$ can be on either side of the line $BN$.

Let's consider the perpendicular bisector of $BN$ (but actually, the diagonal $EF$ is perpendicular to $BN$). Since $BE$ and $EN$ are fixed in length (from the given segments), the vertex $F$ can be constructed on two sides of the line $BN$: one above the line $BN$ and one below the line $BN$. However, we also need to ensure the side lengths. Wait, actually, given $BE$, $EN$ and $BN$, the triangle $BEN$ is fixed (since we know two sides and the included side? Wait, no, $BE$, $EN$ and $BN$: if we consider triangle $BEN$, $BN$ is a side, $BE$ and $EN$ are the other two sides. Now, for the kite $BENF$, $BF = BE$ and $NF=EN$ (or $BF = EN$ and $NF = BE$, but let's check the first case).

If we fix $B$, $E$, $N$ with $BN$ as a base, the point $F$ must be such that $BF = BE$ and $NF=EN$. The set of points $F$ such that $BF = BE$ lies on a circle centered at $B$ with radius $BE$, and the set of points $F$ such that $NF = EN$ lies on a circle centered at $N$ with radius $EN$. The intersection of these two circles (excluding point $E$) will give the possible positions of $F$.

Since the two circles (centered at $B$ and $N$ with radii $BE$ and $EN$ respectively) will intersect at two points (one on each side of the line $BN$), we can form two different kites. Wait, but let's think about the perpendicularity. The diagonal $EF$ is perpendicular to $BN$. For each of the two intersection points of the circles (centered at $B$ and $N$ with radii $BE$ and $EN$), the diagonal $EF$ will be perpendicular to $BN$ (due to the kite's property of diagonals being perpendicular in a kite with two pairs of adjacent equal sides). So, we can have two different kites: one with $F$ on one side of $BN$ and one with $F$ on the other side of $BN$. But wait, is there a case where it's only one? No, because the two circles (unless they are tangent, but here $BE$, $EN$ and $BN$ form a triangle, so the circles intersect at two points). Wait, actually, in the construction, given $BE$, $EN$ and $BN$, the fourth vertex $F$ can be constructed in two ways: above the line $BN$ and below the line $BN$, forming two distinct kites. Wait, no, let's re - examine.

Wait, the problem says "construct kite $BENF$ given sides $\overline{BE}$, $\overline{EN}$ and diagonal $\overline{BN}$". Let's consider the definition of a kite: two pairs of adjacent sides equal. So $BE = BF$ and $EN=FN$, or $BE = FN$ and $EN = BF$. But if we take $BE = BF$ and $EN=FN$, then as $F$ is a point such that $BF = BE$ and $FN=EN$, the two possible positions of $F$ are on either side of the line $BN$. But also, if we consider the other case ($BE = FN$ and $EN = BF$), but in that case, would it be the same? No, because the lengths $BE$ and $EN$ are given, so if $BE
eq EN$, then $BF = EN$ and $FN = BE$ would give a different kite? Wait, no, let's check the given diagram. The diagram shows $B$, $E$, $N$ on a line (the lower line has $B$ and $N$, and the upper line has $B$, $E$ and $E$, $N$). Wait, maybe the key is that the diagonal $BN$ is fixed, and the other diagonal (from $E$ to $F$) is perpendicular to $BN$. So, for the kite, $F$ can be on two sides of the line $BN$ (above and below) such that $BE = BF$ and $EN=FN$ (or vice - versa, but in this case, since $BE$ and $EN$ are given as sides, the two possible kites are formed by having $F$ on either side of the line $BN$). Wait, actually, in a kite, one diagonal is the axis of symmetry. Here, $BN$ is a diagonal, and the axis of symmetry is the line perpendicular to $BN$ passing through the mid - point of $BN$ (or passing through $E$? Wait, $E$ is on the line $BN$ (from the diagram: the upper line has $B$, $E$, $E$, $N$ and the lower line has $B$, $N$). So $E$ is on $BN$. So the diagonal $EF$ is perpendicular to $BN$ (since in a kite, diagonals are perpendicular). So $F$ must lie on the line perpendicular to $BN$ at $E$ (because $E$ is on $BN$). Wait, that's a better way: since $E$ is on $BN$, the diagonal $EF$ is perpendicular to $BN$, so $F$ is on the line perpendicular to $BN$ at $E$. Now, $BF = BE$ and $NF=EN$ (since it's a kite). So $F$ is a point such that $BF = BE$ and $NF = EN$, and $F$ is on the perpendicular to $BN$ at $E$. The set of points $F$ with $BF = BE$ is a circle centered at $B$ with radius $BE$, and the set of points $F$ with $NF=EN$ is a circle centered at $N$ with radius $EN$. The line perpendicular to $BN$ at $E$ will intersect these two circles at two points (one on each side of $BN$)? Wait, no, the line perpendicular to $BN$ at $E$ is a vertical line (if $BN$ is horizontal). The circle centered at $B$ with radius $BE$ and the circle centered at $N$ with radius $EN$ will intersect this perpendicular line at two points? Wait, let's take coordinates. Let’s place $B$ at $(0,0)$, $N$ at $(c,0)$, and $E$ at $(a,0)$ where $0 < a < c$. The line perpendicular to $BN$ (the x - axis) at $E$ is $x = a$. The circle centered at $B(0,0)$ with radius $BE=\sqrt{(a - 0)^2+(0 - 0)^2}=a$ (wait, no, if $B$ is at $(0,0)$, $E$ is at $(a,0)$, then $BE=a$, and the circle centered at $B$ with radius $BE$ is $x^{2}+y^{2}=a^{2}$. The circle centered at $N(c,0)$ with radius $EN = c - a$ (since $E$ is at $(a,0)$ and $N$ is at $(c,0)$) is $(x - c)^{2}+y^{2}=(c - a)^{2}$. The line $x = a$ intersects $x^{2}+y^{2}=a^{2}$ at $(a,y)$ where $a^{2}+y^{2}=a^{2}\Rightarrow y = 0$ (which is point $E$) and that's not good. Wait, maybe my initial assumption about the position of $E$ is wrong. Maybe $E$ is not on $BN$? Wait, the diagram shows two lines: the lower line has $B$ and $N$, the upper line has $B$, $E$ and $E$, $N$. So $E$ is between $B$ and $N$ on the upper line, and $B$ and $N$ are on the lower line. So $BN$ is a horizontal line (lower line), and $BE$ and $EN$ are horizontal segments on the upper line. So $BE$ and $EN$ are horizontal, and $BN$ is horizontal. Wait, that can't be, because then the kite would be degenerate. Wait, no, maybe the diagram is just a sketch. Let's go back to the kite definition: two pairs of adjacent equal sides. Given $BE$, $EN$, and $BN$, we need to find $F$ such that $BENF$ is a kite. So either $BE = BF$ and $EN=FN$ or $BE = FN$ and $EN=BF$.

Case 1: $BE = BF$ and $EN=FN$. The locus of $F$ with $BF = BE$ is a circle centered at $B$ with radius $BE$. The locus of $F$ with $FN = EN$ is a circle centered at $N$ with radius $EN$. These two circles intersect at two points (since $BE$, $EN$, and $BN$ form a triangle, so the circles intersect at two points: one on one side of $BN$ and one on the other side).

Case 2: $BE = FN$ and $EN=BF$. But this would be the same as case 1, just swapping $F$'s position, but actually, no, because the two intersection points of the circles are the two possible positions for $F$, giving two different kites (one above $BN$ and one below $BN$). Wait, but actually, in a kite, the two diagonals are perpendicular. If we consider the diagonal $BN$ and the other diagonal $EF$, when $F$ is above $BN$ and when $F$ is below $BN$, we get two distinct kites. However, wait, let's think about the given sides: $BE$ and $EN$ are given as sides, and $BN$ is a diagonal. So the triangle $BEN$ is fixed (SSS? Wait, $BE$, $EN$, $BN$ are the sides of triangle $BEN$). Now, for the kite, $F$ must be such that $BF = BE$ and $NF=EN$ (so triangle $BEF$ and $NEF$ are isosceles). The two possible positions of $F$ (on either side of the plane relative to $BN$) will give two different kites. But wait, actually, when you construct a kite with given two adjacent sides and a diagonal, there are two possible kites: one on each side of the diagonal $BN$. But wait, no, let's consider the perpendicular bisector. The diagonal that is not $BN$ (i.e., $EF$) must be perpendicular to $BN$ and bisect it (if $BN$ is not the axis of symmetry) or $BN$ is the axis of symmetry. Wait, if $BN$ is the axis of symmetry, then $F$ must be the reflection of $E$ over $BN$, but that would make $BE = BN$ or something, which is not the case. Wait, I think the key is that with the given sides $BE$, $EN$ and diagonal $BN$, we can construct two kites: one where $F$ is on one side of the line containing $BN$ and one where $F$ is on the other side. But actually, in reality, when you have two circles (centered at $B$ and $N$ with radii $BE$ and $EN$) intersecting, they intersect at two points, so two kites are possible. Wait, no, let's take a simple example: let $BE = 1$, $EN = 1$, and $BN = 2$. Then the circles centered at $B$ (radius 1) and $N$ (radius 1) with $BN = 2$ will be tangent at the mid - point of $BN$, but that's a degenerate case. If $BE = 2$, $EN = 3$, $BN = 4$, then the circles will intersect at two points. But in our problem, since $BE$, $EN$, and $BN$ are given as segments (from the diagram, $BE$ and $EN$ are parts of the upper line, and $BN$ is the lower line, so $BE+EN>BN$ (triangle inequality), so the circles intersect at two points. Wait, but the answer is 2? Wait, no, wait: in a kite, one diagonal is the perpendicular bisector of the other. If $BN$ is a diagonal, and $E$ is a vertex, then the other diagonal (connecting $E$ and $F$) must be perpendicular to $BN$ and bisect it. But $E$ is on $BN$ (from the diagram), so the diagonal $EF$ is perpendicular to $BN$ at $E$. So $F$ is on the line perpendicular to $BN$ at $E$. Now, $BF = BE$ and $NF=EN$. So $F$ is a point on the perpendicular to $BN$ at $E$ such that $BF = BE$ and $NF = EN$. The distance from $B$ to $E$ is $BE$, so $F$ is at a distance $BE$ from $B$ and $EN$ from $N$ on the perpendicular to $BN$ at $E$. So we can have two points: one above the line $BN$ and one below the line $BN$ (since the perpendicular line has two directions from $E$). Wait, but if we go above $E$ and below $E$ on the perpendicular, we get two points $F$ such that $BENF$ is a kite. So the number of different kites possible is 2? Wait, no, wait: let's think again. The kite has two pairs of adjacent equal sides: $BE = BF$ and $EN=FN$, or $BE = FN$ and $EN=BF$. But if $BE
eq EN$, then these are two different cases. But in the diagram, $BE$ and $EN$ are on the same line (upper line), so $BE$ and $EN$ are adjacent sides, so $BENF$ must have $BE = BF$ and $EN=FN$ (since $BE$ and $EN$ are adjacent sides of the kite at vertex $E$). So $F$ must satisfy $BF = BE$ and $FN = EN$. The set of points $F$ with $BF = BE$ is a circle, and with $FN = EN$ is another circle. The intersection of these two circles (excluding $E$) gives two points, so two kites. Wait, but the correct answer is 2? Wait, no, actually, when you have a diagonal $BN$ and sides $BE$ and $EN$, the fourth vertex $F$ can be constructed in two ways: one on one side of the line $BN$ and one on the other side, resulting in two different kites.
## Step3: Conclusion
After analyzing the properties of a kite (two pairs of adjacent equal sides, diagonals perpendicular) and the locus of the fourth vertex $F$ (intersection of two circles centered at $B$ and $N$ with radii $BE$ and $EN$ respectively), we find that there are two possible positions for $F$, hence two different kites.

# Answer:
2

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Question

Explanation:

Step1: Recall Kite Properties

Step2: Analyze the Position of \(F\)

Answer: