Trapezoid Riddle II

Let the height of the right trapeziod $AD = h$ and its base $CD=b$ , the area of blue region $[AED] = A_{\color{#3D99F6}blue}$ and that of red region be $[BEC] = A_{\color{#D61F06}red}$ . We note that:

$\begin{aligned} [ACD] & = \frac 12 hb \\ [BCD] & = \frac 12 hb \\ \implies [ACD] & = [BCD] \\ {\color{#3D99F6}[AED]} + [ECD] & = {\color{#D61F06}[BEC]} + [ECD] \\ A_{\color{#3D99F6}blue} & = A_{\color{#D61F06}red} \end{aligned}$

$\implies$ Both are equal.

Including the bottom white triangle, both triangles (white + blue and white + red) share the same base and same height.

Relevant wiki: Crossed Ladders Problem

Let us denote the vertices and lengths as shown above, and $EG // AB // CD$ and passes the intersection point $F$ . According to Crossed Ladders Theorem , $\dfrac{1}{c} =\dfrac{1}{a} + \dfrac{1}{b}$ or $c = \dfrac{ab}{a+b}$ .

Then we are attempting to prove that $c$ is, in fact, half the Harmonic Mean of $a$ and $b$ or, namely, that $EF = FG$ .

By drawing a perpendicular line to $EG$ with intersection at point $H$ , suppose $HG = x$ .

Then by similarity, $\dfrac{b-a}{h_{1}+h_{2}} = \dfrac{x}{h_{1}}$ and $\dfrac{c}{h_{1}} = \dfrac{b}{h_{1}+h_{2}}$ .

Thus, $\dfrac{h_{1}+h_{2}}{h_{1}} = \dfrac{b}{c} = \dfrac{b-a}{x}$ .

$x = \dfrac{c}{b}(b-a) = c - \dfrac{ac}{b} = c - \dfrac{a^2}{a+b}$ .

Now since $a - c = a - \dfrac{ab}{a+b} = \dfrac{a^2}{a+b}$ , $x = c-(a-c) = 2c - a$ .

Thus, $EG = a + x = 2c$ . Hence, $FG = c = EF$ .

As a result, since both $\triangle AEF$ and $\triangle BFG$ share the same height and base length, they have the same area.

Similarly, since both $\triangle EFD$ and $\triangle GFC$ share the same height and base length, they have the same area.

Finally, that concludes that both red and blue triangles have the same area.

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Note : That also proves that the length of $EG$ is the harmonic mean of $a$ and $b$ .

As $\bigtriangleup ACD$ and $\bigtriangleup BDC$ have same altitude and basis: $Area(\bigtriangleup ACD)=Area(\bigtriangleup BDC)$ As blue area is: $Area(\bigtriangleup ACD)-Area(\bigtriangleup EDC)=Area(\bigtriangleup BDC)-Area(\bigtriangleup EDC)$ We find: $Blue Area=Red Area$

You can solve this question by applying Cavalieri's principle. You can create an isosceles trapezoid with same height and base lengths. By symmetry you have that both areas are equal. But also, applying Cavalieri's principle, the triangles of the problem must have the same area of their respective symmetric triangles. Therefore the areas are equal.

Triangle ABD and triangle CBA are at the same base AB and between the same parallels AB and DC. So their areas are equal.

Now, Euclid's axiom states that when equals are subtracted from equals, the results are also equal.

So, area of triangle ABD-area of triangle ABE=area of triangle ABC-area of triangle ABE

Or area of triangle AED=area of triangle BEC

I thought of the right trapeziod as being part of a rectangle with a corner missing. I know that if the red and blue areas were in a rectangle, their areas would be identical. Now say we begin reducing the length of the top line of the rectangle slowly to the left, (forming a right trapezoid). We continue to shorten the top line of the trapezoid until we have a right triangle, at which point both the red and blue areas would disappear simultaneously. I interpolated that neither colored area could be larger than the other anywhere along the way.

You could also solve using similar triangles. notice that triangles AEB and CED are similar.

$\therefore \frac{AE}{EC} =\frac{EB}{ED} = \text{(say)} \alpha. \text{ Now } AE= \alpha EC \text{ and } EB = \alpha ED.$

Therefore it's easy to see that area of triangle AED = area of Triangle BEC (for the area you can the formula use Area= $(1/2) a b \sin(\theta) )$ .

The area of a triangle can be computed as the product of the lengths of two legs and the sin of the angle between them. Taking the center vertex, the opposite angles on the left and right are the same. The correspondingly adjacent legs of the red and blue triangles have equal products since the white triangles are similar.

Triangles A, B, and C are similar right triangles.

Therefore $\frac{s1}{s2}$ = $\frac{s2}{s3}$ = $\frac{s3}{s4}$ Therefore s2 s3 = s1 s4, and the red and blue triangles have the same area.

Here's how I did it: Let $b_1 = \text{length of top base}$ Let $b_2 =\text{length of bottom base}$ Let $h = \text{height of trapezoid}$ Let $B = \text{area of Blue triangle}$ Let $R =\text{area of Red triangle}$ Let $S = \text{area of Small white triangle}$ Let $L = \text{area of Large white triangle}$

I know the area of the bigger right triangle (the one made up of the Blue triangle and Large white triangle) is: $\frac{1}{2}b_2h$

I also know the area of the smaller right triangle (the one made up of the Blue triangle and the Small white triangle) is $\frac{1}{2}b_1h$

Thus, $(1.) B + L = \frac{1}{2}b_2h$ $(2.) B + S = \frac{1}{2}b_1h$

Using the area formula of a trapezoid, I know: $\text{Total Area} = \frac{1}{2}h(b_1 + b_2)$

Using the distributive and commutative properties, I can modify this:

$\text{Total Area} = \frac{1}{2}b_1h + \frac{1}{2}b_2h$

Using substitution from equations (1.) and (2.) above, I can conclude:

$\text{Total Area} = B + L + B + S$ and thus by simplifying the right side:

$(3.) \text{Total Area} = 2B + L + S$

I also know that the Total Area equals the sum of the areas of the 4 triangles, thus:

$\text{Total Area} = B + R + L + S$

Using substitution from equation (3.) above, you get:

$B + R + L + S = 2B + L + S$

So if you subtract "L" and "S" from both sides, you get:

$B + R = 2B$

So now you can subtract "B" from both sides and you get:

$R = B$

This proves that area of the Red triangle is equal to the area of the Blue triangle. :)

Simple,make AB=DC. You now have a rectangle, so the 4 triangles formed by the crossing diagonals are obviously equal. Note: The definition of "right trapezoid" is fuzzy. Either "a trapezoid with two right angles" or "at least two right angles". If one insists on applying the first definition, simply make DC=AB+ delta, with delta approaching zero.