Prove that A rectangle is a square **if and only if** its diagonals are perpendicular.

A rectangle is a square if and only if its diagonals are perpendicular.

I know the proof should have two parts

1. if a rectangle is a square, then its diagonals are perpendicular and
2. if the diagonals in a rectangle are perpendicular, then the rectangle is a square.

But I honestly don't know how to prove them. Can someone help me out?

• Break the rectangle down into the four triangles formed by the intersecting diagonals. Note which if any are congruent. That will depend upon the angle of intersection and the lengths of the diagonals. What conclusions can be reached? – fleablood Apr 16 '17 at 20:12

Hint: an approach to proving "$A$ if and only if $B$" that sometimes works out nicely is to prove that each of $A$ and $B$ is equivalent to some third assertion $C$. In this case, consider rotating the rectangle, $R$ say, through $90^o$ to get a new rectangle, say $R'$. Then the following are equivalent:

• $R$ is a square,
• $R = R'$ (as sets of points in the plane),
• the diagonals of $R$ and $R'$ are perpendicular.

This can be proven algebraically.

Imagine you have the following points on a grid: $(0,0)$, $(0,a)$, $(b,0)$ and $(a,b)$. If $a=b$, these points will form a square. If $a\ne b$, the points will form a rectangle.

Next, take the slopes of the opposite corners. I will call these slopes $m_1$ and $m_2$.

$$m_1=\frac{b}{a}\\ m_2=\frac{-b}{a}$$

Recall that slopes are are perpendicular if and only if they are opposite reciprocals. So, in order for the lines $y=m_1 x$ and $y=m_2 x$ to be perpendicular,

$$m_1 = \frac{-1}{m_2}\\$$

Substitute $m_1\rightarrow\frac{b}{a}$ and $m_2\rightarrow\frac{-b}{a}$

$$\frac{b}{a}=\frac{a}{b}$$

Obviously, the above equality is true only when $a=b$, which, as stated previously, means that the points formed a square. So only squares have perpendicular diagonals.

• Personally, I prefer synthetic proofs. Somehow the algebra masks the simplicity of things. – Alfred Yerger Apr 16 '17 at 22:45

Draw pictures as I write:

Draw the four corners in clockwise order from topleft. The corners are $A$, $B$, $D$ and $C$. (That is in clockwise order. That top from left to right are $A$ and $B$ and the bottom from left to right are $C$, $D$.

Draw the two diagonals. $\overline{AC}$ and $\overline {BC}$ and let them intersect at $E$.

Consider the two triangles $AED$ and $AEB$. Now $AE = AE$ because they are the same line. $ED$ equal $EB$ because diagonals of rectangles bisect each other. [$*$]. So $\triangle AED \cong \triangle AEB$ by $SAS$ if and only if $\angle AED \cong AEB$. As $\angle AED$ and $\angle AEB$ are linear, they are congruent to each other if and only they are both right angles, in other words if and only if the diagonals are perpendicular to each other.

That's it.

[$*$] In a parallelagram the diagonals bisect each other proof.

Use the same point labeling as above. Consider $\triangle AED$ and Triangle $CEB$. $\angle EAD \cong ECB$ and $\angle EDA \cong EBC$ because $\overline{AD}$ is parallel to $\overline {BC}$. And $\overline {AD} \cong {BC}$ because $ABDC$ is a parallelagram. So $\triangle AED \cong CEB$ by $SAS. And so$AE = ED$and$BE = CE\$. So diagonals bisect each other.

One more proof:

You need: 1) In a parallelogram the diagonals bisect each other. 2) In a rectangle the diagonals are of equal length.

You can draw a circle, center M is the point of intersection of the diagonals. Join M to A,B,C and D. There are 4 isosceles triangles with vertex at M, two sides with r = 1/2 × length of diagonal.

Now: 1) Rectangle with perpendicularly intersecting diagonals is a square.

Pick one of the angles at M. It is 90°(diag. intersect perpend.) All the other angles at M are 90°. The 4 triangles emanating from M are congruent by SAS, (Side = r, Angle at M = 90°, Side =r). Hence length AB = length BC = length CD = length DA, ABCD is a square.

Now 2): Given a Square show that the diagonals intersect perpendicularly.

Same drawing as before, circle with center M. The 4 isosceles triangles formed by M have side lengts r, r and ,say, a (a= side length of the square) are congruent by SSS (r,r,a). The corresponding angles are equal. At M, center of the circle, 4 equal angles emanate whose sum is 360°. Hence each is 90°. The diagonals intersect perpendicularly.

Makes sense?