Prove Angle of Diagonals in Quadrilateral is Degrees

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In summary, the conversation discusses a geometric problem involving a square and a strip, with the goal of proving that the diagonals of the quadrilateral formed by the intersecting points of the square and strip intersect at a 45 degree angle. The conversation includes both an algebraic and geometric solution, with the hint that it is enough to note that the triangles formed are isosceles.
  • #1
maxkor
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Two angles of a square with side
c7d457e388298246adb06c587bccd419ea67f7e8.png
protrude beyond a strip of width
c7d457e388298246adb06c587bccd419ea67f7e8.png
with parallel edges. The sides of the square intersect the edges of the strip at four points. Prove that the diagonals of the quadrilateral whose vertices are these points intersect at an angle of
f27da4fca6f3fecb215974023aad210b70ec3857.png
degrees.
rys.png
 

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  • #2
As in a previous problem, here is an algebraic solution. I would like to see a geometric solution giving more insight into why this result holds.
We may as well take $a=1$. Tilt the diagram so that the square becomes the unit square, and suppose that the parallel lines make an angle $\theta$ with the $x$-axis.

[TIKZ][scale=1.25]
\draw [very thick] (0,0) -- (4,0) -- (4,4) -- (0,4) -- cycle ;
\draw [very thick] (-4,0.5) -- (4,6.5) ;
\draw [very thick] (0,-1.5) -- (8,4.5) ;
\coordinate [label=left:$A$] (A) at (0,3.5) ;
\coordinate [label=above:$B$] (B) at (0.667,4) ;
\coordinate [label=right:$C$] (C) at (4,1.5) ;
\coordinate [label=below:$D$] (D) at (2,0) ;
\draw [very thick] (A) -- (C) ;
\draw [very thick] (B) -- (D) ;
\draw (2.5,0.2) node {$\theta$} ;
\draw (1.3,2.6) node {$\phi$} ;
\draw (-0.5,0) node {$(0,0)$} ;
\draw (4.5,0) node {$(1,0)$} ;
\draw (-0.5,4) node {$(0,1)$} ;
\draw (4.5,4) node {$(1,1)$} ; [/TIKZ]
The equations of the parallel lines are then of the form $y = x\tan\theta + c+d$ and $y = x\tan\theta + c-d$ for some constants $c$ and $d$. The condition that the lines form a strip of width $1$ is $2d\cos\theta = 1$. So $2d = \sec\theta$.

The points where the lines cross the sides of the square are then $$ A = (0,c+d),\qquad B = ((1-c-d)\cot\theta,1), \qquad C = (1,\tan\theta + c-d), \qquad D = ((d-c)\cot\theta,0) .$$ The slope of the line $AC$ is therefore $\tan\theta - 2d = \tan\theta - \sec\theta$.

The slope of $BD$ is $\dfrac{-1}{(2d-1)\cot\theta} = \dfrac{-\tan\theta}{\sec\theta-1}$.

It follows from the formula $\tan(\alpha - \beta) = \frac{\tan\alpha -\tan\beta}{1+ \tan\alpha\tan\beta}$ that if $\phi$ is the angle between $AC$ and $BD$ then $$\begin{aligned} \tan\phi = \frac{(\tan\theta - \sec\theta) - \frac{-\tan\theta}{\sec\theta-1}}{1 + (\tan\theta - \sec\theta) \frac{-\tan\theta}{\sec\theta-1}} &= \frac{(\sec\theta - 1)(\tan\theta - \sec\theta) + \tan\theta}{(\sec\theta - 1) - \tan\theta (\tan\theta - \sec\theta)} \\ &= \frac{\sec\theta(\tan\theta - \sec\theta + 1)}{\sec\theta(1 - \sec\theta + \tan\theta)} \\ &= 1. \end{aligned}$$ It follows that $\phi = 45^\circ$.
 
  • #3
Hint
156973.png

156973.png
 
  • #4
1 / α + β = 90
2 / it is enough to note that the triangles
KNP and MLQ (they have two heights of length "a") so they are isosceles
so |∡NMQ| =90 - α/2 and |∡MNP| =90 - β/2
γ=(α + β)/2=45
 

FAQ: Prove Angle of Diagonals in Quadrilateral is Degrees

What is the definition of a quadrilateral?

A quadrilateral is a polygon with four sides and four angles.

How do you prove the angle of diagonals in a quadrilateral is 90 degrees?

To prove that the angle of diagonals in a quadrilateral is 90 degrees, you can use the properties of parallel lines and corresponding angles. By drawing a diagonal in the quadrilateral, you can create two triangles. Using the fact that opposite angles in a parallelogram are congruent, you can show that the opposite angles in each triangle are also congruent. Since the sum of angles in a triangle is 180 degrees, and the opposite angles are congruent, each angle must be 90 degrees.

What is the relationship between the diagonals of a quadrilateral?

The diagonals of a quadrilateral bisect each other. This means that they intersect at their midpoints, dividing each diagonal into two equal parts.

Can a quadrilateral have diagonals that are not perpendicular?

Yes, a quadrilateral can have diagonals that are not perpendicular. This is the case for a kite, where the diagonals are not equal in length and are not perpendicular to each other.

How does the angle of diagonals in a quadrilateral relate to its types?

The angle of diagonals in a quadrilateral can help determine its type. If the angle of diagonals is 90 degrees, the quadrilateral is a rectangle or a square. If the angle is less than 90 degrees, the quadrilateral is a rhombus or a parallelogram. If the angle is greater than 90 degrees, the quadrilateral is a kite or a trapezoid.

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