The image of parallelogram for help

Let's say we have a parallelogram $\text{ABCD}$.

$\triangle \text{ADC}$ and $\triangle \text{BCD}$ are on the same base and between two parallel lines $\text{AB}$ and $\text{CD}$, So, $$ar\triangle \text{ADC}=ar\triangle \text{BCD}$$ Now the things those should be noticed are that:

In $\triangle \text{ADC}$ and $\triangle \text{BCD}$:

$$\text{AD}=\text{BC}$$ $$\text{DC}=\text{DC}$$ $$ar\triangle \text{ADC}=ar\triangle \text{BCD}.$$

Now in two different triangles, two sides are equal and their areas are also equal, so the third side is also equal or $\text{AC}=\text{BD}$. Which make this parallelogram a rectangle.

Isn't it a claim that every parallelogram is a rectangle or a parallelogram does not exist?

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    $\begingroup$ Funny! The trick is in "Two sides are equal and Area is also equal. So, the third side is also equal" ... which has evident counterexamples. $\endgroup$ – Jean Marie Dec 23 '16 at 11:56
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    $\begingroup$ Exactly, construct a line AB and from it's midpoint D, draw a line DE which is not perpendicular to AB. Then ADE and BDE have the same area though $AE \neq BE$. $\endgroup$ – bat_of_doom Dec 23 '16 at 12:01
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    $\begingroup$ @JeanMarie, How can there be counterexamples. Suppose two sides of two different triangles are equal and their area is also equal, then won't we get third side same if we use herone's formula. As herone's formula does not involve angles. $\endgroup$ – Harsh Kumar Dec 23 '16 at 12:12
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    $\begingroup$ If you use Heron's Formula, and let AC be x and BD be y, you will get an equation in x and y. x=y is a valid solution but you are neglecting the solutions that allow for parallelogram without rectangle $\endgroup$ – Swapnil Rustagi Dec 23 '16 at 12:35
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    $\begingroup$ This question provides an excellent counterexample showing why the reasoning it presents is not valid, so it really provides its own answer. The real reasoning apparently used behind the scenes is only mentioned in the comments, some vague supposition that in Heron's formula if two sides and the outcome are fixed, the third side can be solved for uniquely; this is simply not true, and would be obvious if you tried to spell out the argument. But it is hard find fault with an argument that is absent. $\endgroup$ – Marc van Leeuwen Dec 23 '16 at 20:10

Your reasoning is appreciable, but the problem is that it is wrong.

While tempering with sides and area, you forgot about the angles. In this case , it is just the matter of sines and cosines. Let's see how:

Suppose $\angle ADC=\theta$ and $\angle BCD=180-\theta$.

On using trigonometric formula for area of a triangle, you will get that:

$$ar\triangle ADC=\frac{1}{2}\times AD\times DC\times \sin\theta$$

And , $$ar\triangle BCD=\frac{1}{2}\times BC\times CD \times \sin(180-\theta)=\frac{1}{2}\times BC\times CD \times \sin\theta$$

So, $ar\triangle ADC=ar\triangle BCD$

Now, using Cosine Formula:

$$AC^2=AD^2+DC^2-2\times AD\times DC\times \cos \theta$$

And, $$BD^2=BC^2+CD^2-2\times BC\times CD \times \cos (180-\theta)=BC^2+CD^2+2BC\times CD\times \cos\theta$$

This is enough to show that despite of having same area of $\triangle ADC$ and $\triangle BCD$, We can not say that $AC$ is equal to $BD$, these two are equal only when $\theta=90$ which is obviously the case of a rectangle.


After coming back on the site I saw that this question has got much attention and your comment made me realise that I should improve my post.

You discussed that you came to the result In two different triangles, two sides are equal and their Area is also equal. So, the third side is also equal. by Heron's Formula. Let's see when we approach this problem with Heron's Formula.

For area of a given triangle, We have $\triangle=\sqrt{s(s-a)(s-b)(s-c)}$ where terms have their usual meanings.



$$-16\triangle^2=c^4-2c^2\times (a^2+b^2)+(a^2-b^2)^2$$

$$c^4-2c^2\times (a^2+b^2)+[(a^2-b^2)^2+16\triangle^2]=0$$

Notice that this a polynomial of fourth degree and it is not necessary that all four values of $c$ are equal. The answer of your question comes exactly from here.

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    $\begingroup$ Not much of a deal but it is Heron's or Hero's formula not Herone's formula. $\endgroup$ – A---B Dec 25 '16 at 10:08

If you know two sides and the area of a triangle, there will generally be two different lengths for the third side that gives you that area.

Consider, for example: If the two known sides are $3$ and $4$, then the third side is somewhere between $1$ and $7$. A third side of length $1$ gives area $0$, but so does a third side of length $7$. In between, as the third side increases from $1$, the area of the triangle will first increase until it reaches a maximum of $6$ (when the third side is $5$), but then the area decreases towards $0$.

Thus every area between $0$ and $6$ will be hit by some third-side between $1$ and $5$, and by some third-side between $5$ and $7$.

If you try to use Heron's formula to derive the third side, you will end up with quadratic equation with two solutions.

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  • $\begingroup$ When a circle of r=BC around C intersect the line AB, the triangles made of the intersections, D and C are your shown possible two triangles. $\endgroup$ – Takahiro Waki Dec 23 '16 at 17:56

Here is the problem:

"Two sides are equal and Area is also equal. So, the third side is also equal"

Take as an example:

Suppose $\angle ADC=60°$ and $\angle BCD =120°$.

You keep getting: $AD=BC$, $DC=DC$ and $S(ADC)=S(BCD)$ but $\Delta ADC \ne \Delta BCD$ because they don't fit the rule (side,angle,side):

$AD=BC$, $DC=DC$ but $\angle ADC \ne \angle BCD$.

P.S: Having the same area is a consequence of both triangle have the same base ($CD$) and the same height (because $AB$ is parallel to $CD$) but that doesn't garantee congruence. In order to see that just move the side $AB$ (keeping it parallel to $CD$ and on the same height) and clearly you will change the angles $\angle ADC$ and $\angle BCD$ but the area will keep the same.


This post is getting some much attention that I think there is something more to say.

enter image description here In the geometric construction above we have the line $t$ parallel to $s$. Those two paralelogram are just examples about what is happening. For any choice of $AB$ we will always have $AD=BC$ and $DC=DC$ and once $s$ and $t$ are parallels then the height $h$ is constant. It means that we will always have:

$$ar(ADC)=ar(BCD)=\frac{m\cdot h}{2}$$

and clearly there are infinite angles $\angle ADC$ that are not $90º$ and the parallelogram will not be a rectangle. Furthermore if $\angle ADC \ne 90º$ then $\angle ADC \ne \angle BCD$ and once the rule (side,angle,side) can be the definition of congruence between two triangles we will get $BD \ne AC$.

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  • $\begingroup$ Well done on simply and directly addressing the error in the fake proof. This answer deserves more attention. Simplicity seems much undervalued.... $\endgroup$ – Wildcard Dec 23 '16 at 23:21
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    $\begingroup$ This edit has made the post great. +1 $\endgroup$ – Vidyanshu Mishra Dec 24 '16 at 14:15
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    $\begingroup$ Thank you @Lone. I was thinking about a more geometric explanation and I realized that maybe a picture would help people understand what I was trying to say. That's why the EDIT. $\endgroup$ – Arnaldo Dec 24 '16 at 14:30

This is a completion to the answer by TheLoneWolf.

Lemma Let $ABC$ and $DEF$ be two triangles such that $AB=DE$ and $AC=DF$. Then the two triangles have the same area if and only if $\angle A=\angle D$ or $\angle A+\angle D=180^\circ$.


$$\mbox{Area}(ABC)= \mbox{Area}(DEF) \Leftrightarrow \\ \frac{AB \cdot AC \cdot \sin(A)}{2} =\frac{DE \cdot DF \cdot \sin(D)}{2}\Leftrightarrow \\ \sin(A) =\sin(D)\Leftrightarrow \sin(A) -\sin(D)=0 \Leftrightarrow \\ 2 \cos(\frac{A+D}{2})\sin(\frac{A-D}{2})=0\Leftrightarrow \\ \frac{A+D}{2}=90^\circ \mbox{ or } \frac{A-D}{2}=0^\circ$$ where the last line follows from the fact that all angles are between $0^\circ$ and $180^\circ$.

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As already hinted in comments and answers, you appear to have applied Heron's formula to the areas of the triangles $\triangle ADC$ and $\triangle BCD$. Then observing that the only variables in Heron's formula are the three sides of the triangle and its area, and that three of these quantities in the formula for $\triangle ADC$ are equal to the corresponding quantities in the formula for $\triangle BCD$, you concluded that the fourth quantity must also be the same.

I will try to show that Heron's formula not only allows for multiple solutions but (in the case of a non-rectangular parallelogram) gives exactly the two distinct lengths of the diagonals.

One usually sees Heron's formula in a symmetric form such as this: \begin{align} s &= \tfrac12(a+b+c),\\ \triangle &= \sqrt{s(s-a)(s-b)(s-c)}. \end{align} This format is slightly deceptive since the second line shows only one explicit occurrence of the variable $c,$ but there are four implicit occurrences of $c$, one in each place where $s$ occurs. A more explicit formula is $$ \triangle = \tfrac14 \sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}. $$

Algebra shows that the quantity inside the radical can be regrouped as follows: $$ \triangle = \tfrac14 \sqrt{(2ab)^2 - (a^2+b^2-c^2)^2}. \tag1 $$ Now, given $\triangle = \text{Area}(\triangle ADC) = \text{Area}(\triangle BCD),$ $a = AD = BC,$ and $b = CD,$ suppose that $c$ is a solution to Equation $1.$ In order for the quantity under the radical in Equation $1$ to be positive, it must be true that $c^2\leq 2(a^2+b^2).$ Let $$ c' = \sqrt{2(a^2+b^2) - c^2}. \tag2 $$ Squaring both sides of Equation $2$ and rearranging terms, $$ a^2 + b^2 - c'^2 = -(a^2+b^2-c^2) \tag3 $$ and the squares of both sides of Equation $3$ are equal. Hence we can substitute $(a^2 + b^2 - c'^2)^2$ for $(a^2 + b^2 - c^2)^2$ in Equation $1$, that is, $c'$ is also a solution of that equation.

It is still possible that $c'$ is the same as $c.$ This happens precisely when $c^2 = a^2 + b^2,$ that is, when $\triangle ADC$ and $\triangle BCD$ are right triangles. In that case the parallelogram is a rectangle, its diagonals are equal, and the two triangles are congruent. But any value of $c$ such that $\lvert a - b \rvert < c < a+b$ may be a diagonal of a parallelogram with sides $a$ and $b,$ and for every such value such that $c \neq \sqrt{a^2 + b^2}$ there is a distinct value $c'$ that is the length of the other diagonal.

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    $\begingroup$ +1 I believe this most directly addresses his confusion. Heron's formula does not show that if two triangles have two sides with the same lengths and both have the same area, then the 3rd side must be the same. Instead, it only shows that the squares of the 3rd sides must both be solutions of the same quadratic equation. But in general quadratics have two solutions. $\endgroup$ – Paul Sinclair Dec 23 '16 at 15:28

I want to give a more analytic (and probably less intuitive) way of seeing why this is not true. Let's fix $A$, the area of a triangle. Let $a,b>0$ be the respective lengths of two sides of a triangle. Now, let $x$ be the angle between the side of length $a$ and the one of length $b$.

Then, we know that the area is given by:

$$A=\frac{ab\sin x}{2}$$

which yields $\sin x=\tfrac{2A}{ab}$. Now, we also now that the third side has length $c(x)$:

$$c(x)=\sqrt{a^{2}+b^{2}-2ab\cos x}$$

The claim of the OP is that $c:(0,\pi)\to\Bbb R_{0}^{+}:x\mapsto c(x)$ is a constant function. Now, just note that if $\pi > x \geq \pi/2$, then $-1<\cos x\le 0$ and it that case

$$\cos x=\color{red}{-}\sqrt{1-\sin^{2}x}=-\sqrt{1-\frac{4A^{2}}{a^{2}b^{2}}}=-\sqrt{\frac{a^{2}b^{2}-4A^{2}}{a^{2}b^{2}}}=-\frac{\sqrt{a^{2}b^{2}-4A^{2}}}{ab}$$

by the fundamental trigonometric identity $\sin^{2}x+\cos^{2}x=1$. This yields:


On the other hand, if $0<x\le\pi/2$, then $0\le\cos x<1$ and:

$$\cos x = \color{red}{+}\frac{\sqrt{a^{2}b^{2}-4A^{2}}}{ab}$$

which yields


We see that given a certain area $A$ and two side lengths $a$ and $b$, there are only two possible values of $c$, depending on $x$ being an obtuse or an acute angle. We also see that if the two triangles are rectangle, then their respective third sides must be equal, but this already follows from Pythagora's theorem.

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Area of a triangle is

$$ Area = \frac12 b\, c \sin A $$

Among the three, two parameter values excepting $A$ are given, then the remaining angle $A$ and its supplement can satisfy this relation.

If you take base $b$ and $ h= c \sin A $ as height all the following shaded areas have same magnitude:

When area and height remain same (red line is height) $\sin A $ is same. Taking inverse of sine we can have two solutions $\alpha,\beta$ which are supplementary :

$$\alpha+\beta = \pi . $$

The above is general solution of two values.

$$ A= \sin^{-1}{ \frac{2 Area}{bc} } , \quad \pi - \sin^{-1}{ \frac{2 Area}{bc} }$$

This is general situation for a parallelogram.

However, only when

$$\alpha= \beta $$

do we have

$$\alpha= \beta = \pi/2 $$

which is situation of right angles for particular case of rectangle.

ParalloArea Same Area

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    $\begingroup$ How does this answer the parallelogram/rectangle question? $\endgroup$ – Tim Dec 24 '16 at 17:09
  • $\begingroup$ Effective answer (+1). I added a (hopefully) more clear sketch to sustain your your description. $\endgroup$ – G Cab Jan 26 '17 at 15:28

A rectangle, by definition has $4$ right angles (the clue is in rect and angle). So, no, no need for all the clever workings. All rectangles are parallelograms, but not all parallelograms can be rectangles. Only the ones with all four corners being $90$ degrees each.

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The answer by Narasimham is fully right.
To visualize it better, consider to flip $\triangle BCD$ around the bisector of the common segment $DC$.


The resulting sketch shows that angles in $D$ (for $\triangle ADC$) and $C$ (for $\triangle BCD$) are supplementary. They are equal only when they are rect, i.e. when the quadrilateral is a rectangle.

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