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For questions about general quadrilaterals (including parallelograms, trapezoids, rhombi) and their properties.

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Does the square or the circle have the greater perimeter? A surprisingly hard problem for high schoolers

An exam for high school students had the following problem: Let the point $E$ be the midpoint of the line segment $AD$ on the square $ABCD$. Then let a circle be determined by the points $E$, $B$ and ...
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Area of a square inside a square created by connecting point-opposite midpoint

Square $ABCD$ has area $1cm^2$ and sides of $1cm$ each. $H, F, E, G$ are the midpoints of sides $AD, DC, CB, BA$ respectively. What will the area of the square formed in the middle be? I know that ...
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Area of parallelogram = Area of square. Shear transform

Below the parallelogram is obtained from square by stretching the top side while fixing the bottom. Since area of parallelogram is base times height, both square and parallelogram have the same area. ...
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The smallest parallelogram that contains a convex quadrilateral

I try to find the smallest parallelogram in terms of area that contains a convex quadrilateral(A,B,C,D). I am pretty sure it must be constructed from two neighboring sides of the quadrilateral. But ...
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How to find the shaded area

How to find the shaded area crossed by semi-circle of radius 2 and quarter-circle of radius 4?
230 views

Olympiad level | Similar Triangles

The bisector of angle $BAD$ in parallelogram $ABCD$ intersects the lines $BC$ and $CD$ at the points $K$ and $L$ respectively. Prove that the center of the circle passing through the points $C$, $K$, ...
273 views

Arbelos and its angle bisector

I have recently been reading about a very interesting geometry problem and have tried to solve it. I'm now in a point, in which I don't know how to move forward and would appreciate if someone could ...
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Where do you see cyclic quadrilaterals in real life?

I've just been studying cyclic quads in geometry at school and I'm thinking see seems pretty interesting, but where would I actually find these in the real world? They seem pretty useless to me...
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Given a trapezoid with base $AD$ larger than side $CD$. The bisector of $\angle D$ meets $AB$ at $K$. Prove $AK > KB$

We have a trapezoid $ABCD$ with base $AD$ larger than side $CD$. The bisector of $\angle D$ intersects side $AB$ at point $K$. Prove that $AK>KB$. All that I have tried was to make such drawing in ...
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The hunting of “missing primes”

First, I would like to introduce a peculiar way to display the prime numbers (greater than $9$) by means of the ten they belong to ($x$-axis), and their ending digit ($y$-axis). Here's an example of ...
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Maximize area of a quadrilateral given three sides

What is the maximum possible area that a quadrilateral can have, if the lengths of three of its sides are given as 3, 4 and 5, while the fourth side can have arbitrary length? (Thinking of it as three ...
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Calculating area of quadrilateral when distance of vertices from an arbitrary point is known

Given a convex quadrilateral $ABCD$ circumscribed about a circle of diameter $1$. Inside $ABCD$ there is a point $M$ such that $MA^2 + MB^2 +MC^2 + MD^2 =2$. Find the area of the quadrilateral. My ...
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What is the quadrilateral formed by the angle bisectors of a parallelogram?

I have drawn a few parallelograms and their angle bisectors in Geometer's Sketchpad. The quadrilateral looks to me to be a rectangle but how can I prove it ?
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Does a point quadrilateral form a rect in 3D space?

I have 4 points with x and y coordinate and would like to find out a way to check if given quadrilateral would be a rectangle in 3D space. I tried a bunch of conditions, but there was always and edge ...
537 views

How can $4$ points in the plane be vertices of $3$ different quadrilaterals?

Four points on the plane are vertices of three different quadrilaterals. How can this happen? The problem is taken from "Kiselev's Geometry - Book I : Planimetry" At first, I thought it could be ...
568 views

A line segment with a length of 24 makes a 90-degree angle with one of the legs of an isosceles trapezoid. What is the area of this Trapezoid?

Given that $ABCD$ is an isosceles trapezoid and that $|EB|=24$, $|EC|=26$, and m(EBC)=$90^o$. Find $A(ABCD)= ?$ From the pythagorean theorem, I can find that $|BC|=|AD|=10$. Then, I can find the area ...
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Why are parallelograms defined as quadrilaterals? What term would encompass polygons with greater than two parallel pairs?

It seems the definition of a parallelogram is locked to quadrilaterals for some reason. Is there a reason for this? Why couldn't a parallelogram (given the way the word seems rather than as a ...
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Solve for $x$ from this quadrilateral without using law of cosines.

Solve for $x$. I would be able to solve this with law of cosines (with a LOT of work!) but the students that this problem was presented are not familiar with law of cosines. Is there something I am ...
211 views

If a quadrilateral has a pair of equal opposite sides, and a pair of equal opposite angles, then is it necessarily a parallelogram?

I’m sorry I couldn’t upload a photo, so I’ll try to explain it as best as I can. The quadrilateral has a pair of opposite and equal sides, and has a pair of opposite equal angles (85 degrees in the ...
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Show that three circles are coaxal

Let $A_1, A_2, A_3, A_4$ are collinear, $B_1, B_2, B_3, B_4$ are collinear. Such that $A_1, A_2, B_2, B_1$ lie on circle $(O_1)$, and $A_3, A_4, B_4, B_3$ lie on circle $(O_2)$. Let $MNPQ$ be the ...
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Common meeting point for 3 points to reach 4th point [closed]

Problem statement: We are 3 friends at 3 different locations $A, B, C$ and want to reach a location $D$. Each person will take a separate cab to a common meeting point $E$, and then take a single cab ...
194 views

Testing whether the circumcenter of a cyclic quadrilateral lies inside it

For a triangle with sides $a, b, c$ (where $c$ is the biggest side) there is a simple check to see whether it's circumcenter lies inside of it: $$a^2 + b^2 < c^2$$ Is there such an inequality for ...
$ABCD$ is a quadrilateral and $P,Q$ are midpoints of $CD, AB.$…
$ABCD$ is a quadrilateral and $P,Q$ are midpoints of $CD, AB.$ $AP$ and $DQ$ meet at $X, BP$ and $CQ$ meet at $Y.$ Prove that $$|ADX|+|BCY|=|PXQY|$$ (here $|N|$ means area of the shape $N$) I have ...