In-depth explanation on Cyclic Quadrilaterals and Ptolemy's Theorem |
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- Опубликовано: 29 сен 2024
- Learn how to find the diagonal length AB in a quadrilateral. Learn geometry skills: Pythagorean Theorem; Cyclic Quadrilateral; Thales' Theorem; Ptolemy's Theorem. Step-by-step tutorial by PreMath.com
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In-depth explanation on Cyclic Quadrilaterals and Ptolemy's Theorem | #math #maths
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Right! Very Golden! ...And All Rectangles are Cyclic Quadrilaterals, and Squares. I wish ADC were an Equilateral Triangle, then AB = DB + BC. 🙂
Thanks ❤️
As suggested by the title, ACBD is a cyclic quadrilateral. A circle can be drawn so that all four vertices of ACBD lie on the circumference. In this case, we have two 90° angles opposite each other at ∠A and ∠B, so by Thales' Theorem, when inscribed in the circle, DC will become the diameter of the circle.
By Ptolemy's Theorem:
AB•DC = BD•AC + CB•DA
As AC = DA, let AC = DA = z, so this becomes:
AB•DC = xz + yz = (x+y)z = 88z ---> [1]
Triangle ∆DAC:
AC² + DA² = CD²
z² + z² = CD²
CD² = 2z²
CD = √(2z²) = √2z
AB(√2z) = 88z
The problem statement implies that length AB will be the same for any pair of values for x and y that sum to 88. So, let's consider the special case x = y = 44. ACBD can then be proved to be a square. The diagonal of a square is √2 times the length of a side, so AB = 44√2 units for this special case, and, by implication of the problem statement, the same for all pairs x, y where x + y = 88. This is a good answer for a multiple choice test or a test where you are not required to show your work. If you must prove that AB is the same for all values of x and y that sum to 88, knowing the correct answer allows you to check your work for errors.
You are just considering the special case! There is nothing wrong with that.
Thanks ❤️
@@PreMath Yes, this is a special case that is included in the general case. Using different wording from my comment, if we are told that there is only one solution, then the special case's solution must be the same as the general case's solution. When we solve the general case, we can check our work by verifying that this special case solution matches.
Another special case is to let x be infinitesimal and let 88-y also be infinitesimal. x is the distance between B and D and is very small, so BC and DC are almost the same length, and AD and AB are also almost the same length. ΔACD is an isosceles right triangle. With CD having a length extremely close to 88, sides AC and AD are extremely close to 88/√2 = 44√2 and AB, being almost equal to AD, is very close to 44√2, if not exactly 44√2. If we let B and D be the same point, then AB = 44√2. When we prove the general case, we find that AB is, indeed, exactly 44√2 even when B and D are very close but not the same point.
Assume CD = 2r, angle DCB = t, then we have x + y = 2r*(cos t + sin t) = 88.
Then in triangle ABC,
BC = 2r*cos t
AC = √2 r
Angle ACB = π/4 + t
Use cosine formula to get
AB^2 = 2*r^2 + 4*r^2*cos^2 t - 4√2*r^2*cos t * cos (π÷4 + t)
Simplifies to
AB = √2*r*(cos t + sin t) = 88/√2
Thanks ❤️
Very nice! After a long struggle got a similar solution, only much more circuitous, not nearly as elegant as yours.
How can we prove that quadrilateral ACBD is a Cyclic Quadrilateral?
I have the same question
Any triangle is always inscribed in a circle. So let's have a look on the right triangle with sides x and y . Based on the theorem stating that angle at the circumference is half of the angle at the centre that sustains the same arc, if we have 90 degrees as angle at the circumference , then its angle at the centre is 180 , which is a straight line . The hypotenuse is actually the straight line. This means the hypotenuse is the diameter .
Using the same reflexion, the same triangle with equal sides will be inscribed in a circle with the same diameter , since they share the same hypotenuse. As a consequence , the two circles in which the 2 triangles are inscribed will share one diameter , and obviously the same center . They are necessarily the same circle in which the 2 triangles are inscribed
But how can you know if your figure is a cyclic quadrilateral or not ?
My doubt? Isn't any quadrilateral with opposite angles 90⁰, a rectangle? Also, since its two adjacent sides are equal, the quadrilateral has to be a square. The sum of two sides is 88. So, the side is 44. The diagonals should be equal and equal to 44_/2. Am I right?
Hello, actually the other two angles have to add to 180 degrees but don't have to be 90 degrees each
My humble way of solving this problem.
1) If : x + y = 88
2) If : AC = AD
3) If : CD = AB, because CD^2 = AD^2 + AC^2 and CD^2 = x^2 + y^2;
one must conclude that : AD^2 + AC^2 = x^2 + y^2. If so, AD + AC = x + y, and if x + y = 88 AD + AC = 88. As AD = AC; AD = AC = 88/2 = 44.
4) Then : AD + AC must be equal to 88.
5) AD = 44 and AC = 44
6) CD^2 = 44^2 + 44^2
7) CD^2 = 1.936 + 1.936 = 3.872
8) CD = sqrt(3.872) = 62,225
9) As AB = CD , then AB = 62,225 lin un
Answer: AB = 62,225 linear units
You can also solve the puzzle by starting with a square of side x+y. Then in each side find a point with length x and rotate clockwise.if you connect all the points you create an inner square with side √(x^2+y^2). Now if you connect the center of the square with the inner points you create 4 congruent quadrilaterals
Wow! I learned something new, the Ptolemy's Theorem
Dopo vari calcoli di teorema dei seni e coseno,risulta 2(AB)^2=(x+y)^2=>(AB)^2=88^2/2=>AB=88/√2
Excellent!
Thanks ❤️
asnwer=15cm isit
Care about the area? It's (1/4)*(x+y)^2
Good night sir
Same to you dear🌹
Thanks ❤️
Draw Diagonal CD. It is the hypotenuse of the new formed triangles △CAD and △CBD.
Label AC = z. Then AD = z because the segments are congruent.
So, CD = x² + y² = z√2 by the Pythagorean Theorem.
The two opposite angles ∠CAD and ∠CBD are supplementary because they are both right angles.
Then, by the Polygon Interior Angles Theorem, m∠ACB + m∠ADB = 180°. Thus, ∠ACB and ∠ADB are supplementary as well. ACBD is a cyclic quadrilateral.
First time using Ptolemy's Theorem. Let's give it a shot!
AB * CD = (AC * BD) + (AD * BC)
AB * z√2 = zx + zy
= z(x + y)
= 88z
AB = 88/(√2)
= (88√2)/2
= 44√2
So, the diagonal length AB = 44√2 units (exact), or about 62.23 units (approximation).
The result depending only of x +y when B moves on the circle of diameter [C,D] we choose for example x = 88 and y = 0 (meaning that B = C),.
Then AB = AC = DC/sqrt(2) = DB/sqrt(2) = x/sqrt(2) = (x + y)/sqrt(2) = 88/sqrt(2) = 44.sqrt(2).
Thank you for the introduction to Ptolemy's Theorem and cyclic quadrilaterals. As always, I appreciate your educational effort!
44x(2^(1/2))
is there another method to solve inside the diagramm pls answer
Many approaches are possible to find the solution to this problem! You may try Trigonometry. Cheers🌹
At most, x = 88 and B = C or y = 88 and B = D, then AB = 88/√2 = 44√2.
Thanks ❤️