# How to prove that perpendicular from right angled vertex to the hypotenuse is at most half the length of hypotenuse of a right triangle?

In a right-angled triangle $$\Delta ABC$$, prove that the perpendicular $$BD$$, drawn from the right-angled vertex $$B$$ to the hypotenuse $$AC$$, is at most half the hypotenuse $$AC$$.

My approach:

Assume that $$AB=x$$, $$BC=y$$, $$AC=k$$ where $$k$$ is some arbitrary constant

I used Pythagoras theorem in $$\Delta ABC$$

$$y^2=k^2-x^2,\ y=\sqrt{k^2-x^2}$$

I used formula of area of right triangle ABC by two methods & equate them $$\frac12(BD)\cdot(AC)=\frac12x\cdot y\implies BD=\frac{xy}{k}$$ $$BD=\frac{x\sqrt{k^2-x^2}}{k}$$ I differentiated $$BD$$ with respect to $$x$$ $$\frac{d}{dx}BD=\frac{k^2-2x^2}{\sqrt{k^2-x^2}}$$ putting $$d(BD)/dx=0$$, I got $$x=k/\sqrt2$$ & $$y=k/\sqrt2$$

The maximum length of altitude BD will be $$\frac{xy}{k}=\frac{(k/\sqrt2)\cdot(k/\sqrt2)}{k}=\frac k2$$ Above value proves that maximum value of $$BD$$ is half the hypotenuse AC. It is fine but I don't want to use this lengthy proof by calculus .

My question:

Is there any simple or easy proof by using trigonometry, geometry, or other way?

• Thank you for working this out in detail. Commented Jun 17, 2020 at 5:47
• OP that's very interesting you used calculus to solve this - that wasn't even on my radar! I first thought about making assumptions of making the right triangle into an isosceles right triangle. That assumption makes it where it is impossible for the altitude to exceed half the hypotenuse. However, I wanted to not make that assumption yet, and it turns out that there is an Euler proof called the Geometric Mean Theorem already. It's on Wikipedia. I used the Pythagorean Theorem method, but you can see from the page that there are several other ways. Commented Jun 18, 2020 at 7:14
• @G.Khanna Thanks. this was only I could think of.
– user766881
Commented Jun 19, 2020 at 1:29

It can be easily proved by geometry

Consider a right $$\Delta ABC$$ having hypotenuse $$AC$$ of constant length which is inscribed in a semi-circle with center $$O$$ & radius $$OA=OE=OC(=AC/2)$$ (as shown in figure below).

The right angled vertex $$B$$ can lie anywhere on the semi-circular arc AC such that $$BD\le EO$$ $$BD\le \color{blue}{\frac{AC}{2}}$$

• @user1551 E is farthest from the diameter AC of semicircle. Isn't it true? End-points of a diameter are always farthest from the center of any circle. Commented Jun 17, 2020 at 17:40
• I am not saying that it's false. I'm saying that this claim deserves a proof. It can be easily proved by using Pythagorean's theorem, but your argument is not as simple as you might think. Commented Jun 17, 2020 at 17:43
• See, postulates are taken without proofs. 'four triangles have to fit in the (larger) square' in the answer by David Z might also need a proof. But facts/postulates are taken to be true. However if you can prove it mathematically or logically, please do share it. Commented Jun 17, 2020 at 17:49
• Oh, yes, sorry to be unclear, my comment was more directed at @user1551 than you. I meant to say that I agree that the answers need not be fully rigorous. Commented Jun 17, 2020 at 19:02
• @user1551 You can check my second answer not much different but it gives the proof you asked me in comment at the top. Commented Jun 18, 2020 at 19:39

While it uses the same mathematics as some of the other proofs/justifications presented here, I think this visualization is rather intuitive. Make the following square out of four copies of the triangle, plus a smaller square:

The area of the triangle is $$A = \frac{1}{2}\mathrm{AC}\times\mathrm{BD}$$. Since four triangles have to fit in the (larger) square: \begin{align} 4\times\frac{1}{2}\mathrm{AC}\times\mathrm{BD} &\le (\mathrm{AC})^2 \\ \mathrm{BD} &\le \frac{1}{2}\mathrm{AC} \end{align}

This is not a rigorous proof, of course, and I wouldn't even go so far as to call it an intuitive (i.e. non-rigorous but easily understandable) "proof". If you wanted to make it more rigorous, you would need to show that the triangles won't overlap, perhaps using something like the semicircle argument in Harish Chandra Rajpoot's answer.

• You fit four triangles in larger square but why don't you fit small square also to get larger square?
– user766881
Commented Jun 17, 2020 at 18:10
• @Jyerkimenon Unfortunately I don't understand what you mean by that comment. Commented Jun 17, 2020 at 18:22
• i am asking about small square inside. isn't small square a part of larger square? why only four triangles alone make larger square?
– user766881
Commented Jun 17, 2020 at 18:32
• Sure, I suppose you could put it that way. But the important fact is simply that the four triangles fit in the large square, which implies that the total area of the four triangles is no greater than the area of the large square. Commented Jun 17, 2020 at 18:48
• @Cireo Yeah, I was considering that but I decided to skip it because it detracts from the simplicity of the visualization IMO. That's why I'm not presenting this as a proof. But maybe it'd be worth it to edit in a note about that; perhaps I'll do that later. Commented Jun 18, 2020 at 19:02

Given a right triangle with legs $$a$$ and $$b$$, the hypotenuse, $$c$$ satisfies $$c^2=a^2+b^2\tag1$$ Similar triangles show that $$h$$, the altitude on the hypotenuse, satisfies $$\frac ha=\frac bc\tag2$$ Then $$(2)$$ says $$\frac hc=\frac ac\frac bc\tag3$$ and $$(1)$$ says $$1=\left(\frac ac\right)^2+\left(\frac bc\right)^2\tag4$$ Thus, \begin{align} \frac hc &=\frac ac\frac bc\tag5\\ &\le\frac ac\frac bc+\frac12\left(\frac ac-\frac bc\right)^2\tag6\\ &=\frac12\left(\left(\frac ac\right)^2+\left(\frac bc\right)^2\right)\tag7\\ &=\frac12\tag8 \end{align} Explanation:
$$(5)$$: $$(3)$$
$$(6)$$: squares are non-negative
$$(7)$$: algebra
$$(8)$$: $$(4)$$

with equality when $$a=b$$.

• Thank you for working this out in detail. I was looking for something similar! For those who may have gotten lost in Eqn.(6), I would just add one detail: the quantity $\frac{1}{2} \left( \frac{a}{c} - \frac{b}{c} \right)^2$ has to be positive. Commented Jun 17, 2020 at 5:48
• That was the meaning of "squares are non-negative". (Note that it doesn't need to be positive, just non-negative.)
– robjohn
Commented Jun 17, 2020 at 6:15

I suppose the question is really that $$BD \leq \frac{AC}{2}$$. ["the perpendicular BD drawn from right angled vertex B to the hypotenuse AC is $$\bf {at \, most}$$ half the hypotenuse AC.]

Let $$M$$ be the midpoint of $$AC$$. Then $$BD \leq BM = \frac{AC}{2}$$. The inequality follows from the fact that no other point on $$AC$$ is closer to $$B$$ as is $$D$$; to justify the second equality, just "complete the rectangle."

• In other words, a median cannot be shorter than the altitude from the same vertex. Commented Jun 18, 2020 at 22:24

$$\frac {AC}{2}$$ is the arithmetic mean between AD and DC.

Due to the similar triangles ABD and BDC, BD is the geometric mean between AD and DC:
$$AD:BD :: BD:DC$$

The geometric mean is less than or equal to the arithmetic mean between two numbers. Therefore, BD is less than or equal to $$\frac {AC}{2}$$.

Proof of the last statement:

Let b be the geometric mean between a and c. Then:
$$\frac {a}{b} = \frac {b}{c}$$ $$\frac {b-a}{b} = \frac {c-b}{c}$$ If c > b, then $$(c-b) > (b-a)$$, and therefore b is closer to a than it is to c.

• did you use $\frac{AD+DC}{2}\ge \sqrt{AD* DC}$? how can you use it?
– user766881
Commented Jun 17, 2020 at 18:14
• It is correct that $\frac{AD+DC}{2}\ge \sqrt{AD* DC}$. Note that the left-hand side is the arithmetic mean, and the right hand side is the geometric mean. So that is another way of saying that "the geometric mean is less than or equal to the arithmetic mean." The last section of my post is a simple proof of that theorem. Commented Jun 17, 2020 at 19:06

Apply the inequality $$2xy \le x^2+y^2$$ to obtain

$$BD = \frac{xy}k \le \frac{\frac12(x^2+y^2)}k= \frac{\frac12k^2}k =\frac {AC}2$$

i.e. BD is at most half of the hypotenuse.

• Obligatory remark. Essentially the same proof, based on the AM-GM inequality, also appears on pp.15-17 of Kazarinoff's nice little book Analytic Inequalities. According to Kazarinoff, "Another way of defining $x$ [our $BD$ here] is to say that it is the length of a side of a square whose area is equal to that of a rectangle with sides of lengths $a$ and $b$ [i.e. our $AB$ and $BC$]. This definition comes from ancient Greece and can be found in Euclid's Elements." Commented Jun 19, 2020 at 3:19

Let $$A=(0,0)$$ and $$C=(0,1).$$ Also, let $$B=(x,y).$$ Then, $$0=(x,y)\cdot(x-1,y)=x^2-x+y^2.$$ Since $$D=(x,0)$$ and $$0\le x \le 1,$$ $$|BD|^2=y^2=x-x^2 \le 1/4.$$ (On $$[0,1]$$, the maximum of $$x-x^2$$ occurs at $$x=1/2.$$)

• did you consider hypotenuse AC of unit length?
– user766881
Commented Jun 18, 2020 at 20:15
• well so if maximum occurs at $x=1/2$ then how is $BD=AC/2$?
– user766881
Commented Jun 18, 2020 at 23:33
• In any geometric problem, you're free to choose your unit of length. Commented Jun 18, 2020 at 23:47
• Because $|AC|= 1$ (by my choice of length unit) and $|BD| \le 1/2$ (by taking square roots above). The conversion to Cartesian coordinates was made much easier by choosing a suitable unit of length (and putting the corresponding edge along the X-axis). I also chose where to put the origin. Commented Jun 19, 2020 at 0:05
• I assume that you meant $BD \le AC/2$ rather than $BD=AC/2$ above. Commented Jun 19, 2020 at 0:19

Here's how you might show it using trig rules (and calculus, if you want to avoid using the double-angle formulae).

For right triangle ABC, we'll take AB to be the hypotenuse.

Let's assume that the hypotenuse is length 1 (if it's length $$L$$, we can simply rescale all lengths by dividing by $$L$$). We will let the angle at A be $$\theta$$. Given this, length AC is given by $$AC = \cos \theta$$ Now, we create the perpendicular, which meets AB at D. The hypotenuse of the new right triangle including point A is AC. Now, as CD is the opposite side to point A (which is still $$\theta$$), we can determine that $$CD = AC\sin\theta$$, or $$CD = \cos\theta \sin\theta$$ If you know your trig functions, in particular your double-angle formulae, you can probably see from this point that CD must be no more than $$\frac12$$ (or half of the hypotenuse, as we normalised to make the hypotenuse length 1). This is because $$\sin 2\theta = 2\cos\theta\sin\theta$$, and thus $$CD = \frac12\sin2\theta$$. Given that $$\sin x\leq1$$, we see that $$CD\leq\frac12$$.

But let's do this with calculus, rather than foreknowledge of trig rules. We seek the extrema of the function $$f(\theta)=\cos\theta\sin\theta$$ bounded by $$0<\theta<\frac\pi2$$. To be an extremum, the derivative must be zero, so $$f'(\theta) = \cos^2\theta - \sin^2\theta=0$$, or $$\tan^2\theta = 1$$. Since $$0<\theta<\frac\pi2$$, and $$\tan\theta>0$$ in this range, we get $$\tan\theta=1$$, or $$\theta=\frac\pi4$$.

At this point, $$\cos\frac\pi4=\sin\frac\pi4=\frac1{\sqrt{2}}$$, and thus we obtain $$f(\frac\pi4)=\frac12$$. As $$f(\theta)$$ is continuous, $$f(0)=f(\frac\pi2)=0$$, and $$\theta=\frac\pi4$$ is the only extremum found in the range, we can conclude that it is the maximum. Therefore, the largest value of CD is $$\frac12$$, and we have proven what we set out to prove.

I am posting this answer to fulfill the requirement of proof asked by @user1551 in a comment on my previous answer.

Consider a right $$\Delta ABC$$ inscribed in a semi-circle with center $$O$$ & radius $$OA=OB=OC(=AC/2)$$. Join the right angled vertex $$B$$ to the center $$O$$ (as shown in figure below).

Since length of hypotenuse $$AC$$ is constant hence right angled vertex $$B$$ can lie anywhere on the semi-circular arc AC such that $$BD=OB\sin\theta$$ $$\color{blue}{BD=\frac{AC}{2}\sin\theta}\quad \quad (\forall \ \ 0<\theta<\pi)$$ The maxima will occur at $$\theta=\dfrac{\pi}{2}$$ i.e. when the right angled vertex $$B$$ lies at the highest point $$E$$ from hypotenuse AC.

• No offence. To be honest, I think this makes your proof worse, because it now becomes a strange mix of elementary plane geometry and coordinate geometry. I don't think it has been proved in any high school that these two kinds of geometry are really the same thing. There is an easier proof: by Pythagorean's theorem: $DB=\sqrt{OB^2-OD^2}\le OB=OE$. Anyway, I think that the answer by Quanto (and the similar ones by Timothy Smith or robjohn) is the best. It uses only properties of similar triangles and the AM-GM inequality. No auxilliary geometric construction is needed. That's true simplicity. Commented Jun 19, 2020 at 2:52
• well, that's great. what about my first proof, is there a mix of two? Commented Jun 19, 2020 at 2:59
• It isn't, but my point is that your first proof isn't as simple as it appears, and I think your new answer illustrates my point very well. Anyway, thanks for giving this second proof. Commented Jun 19, 2020 at 3:11

See the picture. I have used an obvious inequality.

Let E be the midpoint of hypotenuse AC, median joining mid-point of hypotenuse is half in length $$BE=\dfrac{AC}2$$

Use Pythagorean theorem in right triangle BDE

$$BD=\sqrt{BE^2-DE^2}\le BE$$ $$BD\le \dfrac{AC}{2}$$