# Pythagorean Theorem - Incomplete proof

The image below is supposed to prove Pythagorean Theorem (without words). However, I see that it is assuming that the square of side $c$ can be inscribed in the square of side $a+b$ and would result in the $4$ triangles of equal area. This fact is only assumed, so would the proof be considered incomplete? • I'd come at it the other way an say that there is no embedded square there are only 4 triangles. Then putting them together makes the square. The triangles form right angles for the same reason gradients of perpendicular lines. – Karl Mar 17 '18 at 17:21
• The complementary angles plus the right angle of the yellow square sum to a straight angle. In this context, a proof without words is intended to provoke an easy thought process to complete it. – quasi Mar 17 '18 at 17:22
• Surely, the main idea is shown in the picture, and many people would say that it is easy to complete the proof from this picture. But it is debatable, whether it can be considered to be "complete" , a quite subjective word in this context anyway. – Peter Mar 17 '18 at 18:24
• Nevertheless, a very elegant variation of a proof of the pythagorean theorem. – Peter Mar 17 '18 at 18:25
• The answer below shows an important part that I missed : It must also be shown that the yellow figure is actually a square. It may be not very difficult to show this (I leave it to everyone to decide him/herself) , but it is also a necessary part to complete the proof. Considering this, I wouldn't call this picture-proof "complete". – Peter Mar 18 '18 at 9:47

Theorem :

In a right triangle with legs $a,b$ and hypothenuse $c$:

$a^2+b^2=c^2.$

Geometric proof:

Consider a square of side length $a+b$ (drawing).

Partition the sides into $a$ and $b$ as shown.

Each of the $4$ corner triangles formed are congruent by $SAS$, i.e. $a,b$, and right angle between.

Hence the quadrilateral formed by connecting the partition points has $4$ equal sides $c$, a rhombus.

The adjacent sides of the inner quadrilateral are perpendicular (why?).

Hence the inner quadrilateral is a square.

Area of outer square of side length $a+b:$

$A:= (a+b)^2.$

Also:

$A= 4(1/2)ab + c^2$, where $(1/2)ab$ is the area of one triangle (why?).

Finally:

$a^2+b^2 = c^2$.

• Thanks for your solution. I am not sure if you have answered the question...The tricky part is that it is not obvious to me that when you partition a line into two parts at each side and you connect the partition point, you get a square. – NoChance Mar 17 '18 at 21:10
• NoChance.Thought the above proves your point. The inscribed quadrilateral has 4 l sides of equal length c. Rhombus. The angle between adjacent sides is 90°. Hence? – Peter Szilas Mar 17 '18 at 21:53
• Got it thanks... – NoChance Mar 18 '18 at 0:12