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

Question

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?

Answer 1

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 $$BD\le EO$$ $$BD\le \color{blue}{\frac{AC}{2}}$$

Answer 2

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."

Answer 3

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 $a=b$.

Answer 4

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.

Answer 5

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.

Answer 6

$\frac {AC}{2}$ 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 $\frac {AC}{2}$.

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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.

Answer 7

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.

Answer 8

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 $x=1/2.$)

Answer 9

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.

Answer 10

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

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

Answer 11

See the picture. I have used an obvious inequality.