Triangle Inequality?

Use Ptolemy's inequality . That is, for all quadrilaterials,

$\displaystyle \sum_{product \ of \ 2 \ opposite \ sides} \ge \sum_{product \ of \ 2 \ diagonals}$

But...

$(4 \cdot 6) + (7 \cdot 5) \not \ge (8 \cdot 9)$

Hence, the quadrilateral doesn't exist.

By Law of Cosines , we have

$9^2 = 4^2+7^2-2(4)(7) \cos A$ $\color{#20A900}\implies$ $A=106.6015496^\circ$

$9^2=5^2+6^2-2(5)(6) \cos C$ $\color{#20A900}\implies$ $C=109.4712206^\circ$

$8^2=4^2+5^2-2(4)(5) \cos B$ $\color{#20A900}\implies$ $B=125.0996322^\circ$

$8^2=6^2+7^2-2(6)(7) \cos D$ $\color{#20A900}\implies$ $D=75.52248781^\circ$

Adding the angles, we have

$A +B+C+D=106.6015496^\circ+109.4712206^\circ+125.0996322^\circ+75.52248781^\circ=416.6948902^\circ$ $\color{#20A900}\implies$ $\color{#69047E}\text{It is not equal to 360.}$

$\color{#3D99F6}\huge\therefore$ $\large\color{#D61F06}\boxed{\text{It is not a quadrilateral.}}$

If this quadrilateral exists; then the sum of the areas of the 8-6-7 triangle and the 8-5-4 triangle must be equal to the sum of the areas of the 9-5-6 triangle and 9-7-4 triangle.
By heron's formula; the above situation means;
$\sqrt{0.5×8.5×3.5×4.5} + \sqrt{10.5×2.5×3.5×4.5}= \sqrt {10×1×5×4} + \sqrt {10×1×6×3}$
$\implies \sqrt{66.9375} + \sqrt{413.4375} = \sqrt {200} + \sqrt{180}$
A quick look at the calculator or a quick approximation in mind and we can conclude that the LHS is roughly 1 more than the RHS.
Hence; this quadrilateral doesnt exist.

• '.' $9^{2} > 5^{2} + 6^{2}$
• .'. the angle oposite to the side with length 9 is obtuse angle
• '.' $9^{2} > 4^{2} + 7^{2}$
• .'. the angle oposite to the side with length 9 is obtuse angle
• '.' $8^{2} > 5^{2} + 4^{2}$
• .'. the angle oposite to the side with length 8 is obtuse angle

• '.' quadrilateral can't have 3 obtuse angles

• .'. the shown sidescan't make a quadrilateral

A simple graphical solution.

Notice that the left (B) and right (D) points lie on the ellipse with the other two points (A and C) as the two foci, c=4.5 and a=5.5. Midpoi t between AC=O. Then 2(DO^2+4.5^2)=5^2+6^2. Easily DO<4. Similarly BO<4. So it is impossible for BD=8.

Let $\angle 45$ be the angle between sides of lengths 4 and 5 (for arbitrary values of 4 and 5).

Naturally we need that $\angle 49 + \angle 59 = \angle 45$ .

By the cosine rule: $7^2 = 4^2 + 9^2 - 2 \cdot 4 \cdot 9 \cos \angle49 \Rightarrow \cos \angle49 = 2/3 \Rightarrow \sin \angle49 = \frac{\sqrt5}3 \\ 6^2 = 5^2 + 9^2 - 2 \cdot 5 \cdot 9 \cos \angle59 \Rightarrow \cos \angle59 = 7/9 \Rightarrow \sin \angle59 = \frac{4\sqrt2}9 \\ 8^2 = 4^2 + 5^2 - 2 \cdot 4 \cdot 5 \cos \angle45 \Rightarrow \cos \angle45 = 23/40 \\ \sin \angle49 \sin \angle59 = \frac{4\sqrt{10}}{27}$

But $\cos (A + B) = \cos A \cos B - \sin A \sin B$ . We already have a problem: Since $\cos \angle49 \cos \angle59$ and $\cos \angle45$ are both rational, $\sin \angle49 \sin \angle59$ would also need to be rational. But $\sin \angle49 \sin \angle59$ is irrational, so we have a problem.

[This is not a complete solution.It is just for discussion doubts and anomalies.]

Assume that the angle formed by length 4 and length side is A.The diagonal of length 8 divides this angle in two parts namely P and Q.

First find $cos \angle A$ by triangle cosine rule whose sides are $4,7,9$ .

Now calculate $cos \angle P$ whose sides are $4,8,5$ and $cos \angle Q$ whose corresponding sides are $8,7,6$ .

Apply the formula $cos \angle A=cos \angle (P+Q)$ .

While calculating these two values will not be equal

It makes sure the construction of this quadrilateral is not possible.

The side whose length is 4 is longer than the one with 6 this isn't possible and hence such a figure can't​ exist.

Using the 9 diagonal as common base, drop a perpendicular (h1) from the apex of the 9/6/5 triangle. Since the base on 6 side must be <6, the base on the 5 side must be >3, and h is therefore <4. Using the same method on the other triangle (9/7/4) it is clear that the h2 <3. The shortest distance between the two apexes will be greater than the sum of the two heights (which will be less than 8, as demonstrated); but the hypotenuses of the small right triangles between the heights and the "8" diagonal are nevertheless <5 and <3 respectively, so that total diagonal must be <8. (If, like me, you didn't know Ptolemy's inequality).

The almost right triangle with sides 5 and 6 would never Have a 9 as the hypotenuse

$9^{2}=4^{2}+7^{2}-56\cos A \rightarrow \cos A =\dfrac{-37}{56} \\ 9^{2}=5^{2}+6^{2}-60\cos C \rightarrow \cos C =\dfrac{-1}{3} \\ 8^{2}=4^{2}+5^{2}-40\cos B \rightarrow \cos B =\dfrac{-23}{40} \\ 8^{2}=7^{2}+6^{2}-84\cos D \rightarrow \cos D =\dfrac{1}{4}$

There would be three obtuse angles. Impossible!

$\arccos \left( -\dfrac{37}{60}\right) + (\arccos \left( -\dfrac{1}{3}\right) + (\arccos \left( -\dfrac{23}{40}\right) + (\arccos \left( \dfrac{1}{4}\right) = 7.6... rad$

Check using pythagorean theroem if the square of the diagonal is equal to the square of the sum of the other sides.

You don't have to calculate anything. 6-7-8 is nearly equilateral, so the 9 will extend way beyond the 8 line. Which it can't do because the 8-4-5 triangle is nearly a straight line. QED.

By inspection, at least one inscribed triangle violates the Triangle Inequality, hence this quadrilateral is impossible.

Simple: because the triangle formed by the sides of 4, 5, and 9 would not exist, since 4+5=9 and for a triangle to exist, the sum of two sides must be greater than the third.

the full Pythagorean theorum:

if (C x C) = (B x B) + (A x A), the triangle is right. if (C x C) is greater than (B + B) + (A x A), the triangle is obtuse. if (C x C) is less than (B x B) + (A x A), the triangle is acute.

Also, the biggest angle in a triangle is opposite the biggest side.

Plugging in the top left triangle, 4 x 4 is less than 8x8 + 9x9, so the top left angle in the center would be acute (if it were real).

In order for the top right side to be obtuse to match, the 5 from the perimeter would need to be greater than the 9 from the center to the top.

Because 2 acute angles can not add up to 180 degrees, the figure has artificial measurements.

Can this be invalidated by proving the triangles represented can't exist as shown?