Prove that $AB\cdot CD\geq 2S$.

Question

Question:

Let $M$ be the centroid of triangle $ABC$. $D$ is a point on the line passing through $A$ and parallel to $BC$ such that $\angle CMD=90^\circ$. Let $S=[AMCD]$. Prove that $AB\cdot CD\geq2S$.

This question was the last question from a mock exam a few days ago. To clarify, this exam is contest-style and for 7-8th graders.

My approach: I immediately noticed that $2S=AC\cdot MD\sin\alpha$, where $\alpha$ is the angle between $AC$ and $MD$. Hence, it suffices to prove that $AB\cdot CD\geq AC\cdot MD$, since $\sin\alpha\leq1$. This is then equivalent to $\dfrac{AB}{AC}\geq\dfrac{DM}{DC}$. However, I am not sure how do approach the problem from here. Any help on solving this using (preferably) synthetic geometry will be appreciated!

Answer 1

Let $N$ be the midpoint of $CD$. Note that $MN=\frac 12 CD$ as $\angle CMD=90^\circ$. Let $MN$ intersect $AB$ at $X$. Note that $$AB \cdot CD = 2AB\cdot MN \ge 2 AB \cdot MN \cdot \sin \angle AXN = 4[AMBN]$$ and it remains to show that $2[AMBN]=[AMCD]$. But this is very easy: we have $$[AMBN]=[ABN]-[ABM]=\frac 12[ABCD]-\frac13 [ABC]$$ and $$[AMCD]=[ABCD]-[ABCM]=[ABCD]-\frac 23 [ABC].$$ Done!

Addendum. There are other possible configurations of points. For instance, it may happen that points $A$ and $D$ lie on different halfplanes determined by $CM$. Similar ideas work: as before we have $AB \cdot CD \ge 4[AMBN]$. Note that $[ABN]=[ABD]−[ANBD]=[ACBN]−[ACB]$, so taking the average and using $[ANBD]=[ACBN]$ we get $$[ABN]=\frac 12 \Big([ABD]−[ANBD]+[ACBN]−[ACB]\Big)=\frac 12 \Big([ABD]−[ACB]\Big).$$ Therefore $$[AMBN]=[ABN]+[AMB]=\frac 12\Big([ABD] - [ACB]\Big) + \frac 13[ACB] = \frac 12 [ABD]-\frac 16 [ACB].$$ and $$[AMCD]=[ACD]-[ACM]=[ABD]-\frac 13[ACB],$$ which shows that $2[AMBN]=[AMCD]$.

Yet another configuration: points $A$ and $D$ lie on the same halfplane determined by $CM$ and segments $CD$, $AM$ intersect. Then we have simply $$AB \cdot CD \ge 2[ACBD] > 2[ACMD].$$

Answer 2

Comment: Particular case: suppose $CM=\frac{BC}2$, We construct right triangle CDC', such that $DC'=AB$ . We have:

$S_{DC"C}= (C'D=AB)\cdot CD$

Now we reflect MC and D about AM , the result is hexagon AG'C'MCD. We have:

$S_{AG'C'MCD}=2S_{CDAM}$

We connect C to C'. In hexagon $AG'C'MCD$ we delete portion DAG'C' and added portion MC'C and get triangle DC'C. Now we have to show that:

$S_{AG'C'D}<S_{MC'C}$

In this case:

$2S_{ADCM}<S_{DC'C}=(DC'=AB)\cdot DC$

In this figure we see:

$S_{AG'CD}\approx S_{CIHD'''M'_1M}<S_{MCC'}$

Answer 3

Solution for isosceles ABC:

I. With $M$ centroid, $AD\parallel BC$, and $DM\perp MC$, let $\triangle ABC$ be isosceles with $AB=AC>BC$, and such that $CD\perp BC$. This occurs when semi-base $CG$ is the mean proportional between $GM$ and $MA$, i.e.$$\frac{GM}{CG}=\frac{CG}{MA}$$whence $$AB>GM+MA=AG>2CG=BC$$Thus, since $AM=2MG$, $ABC$ is the isosceles triangle in which$$\frac{altitide}{base}=\frac{3}{2\sqrt 2}$$From $C$ make $CK=BF$ and complete rectangle $CKND$. Since $AB>BC$, and $F$ and $G$ are midpoints of $AB$ and $BC$, then$$CK>CG=AD$$and$$CKND>ADCM$$And since$$CKND=BF\cdot CD=\frac{1}{2}AB\cdot CD$$then$$AB\cdot CD>2ADCM$$

II. Now, keeping $BC$ fixed, suppose altitude $AG$ increased as in the next figure.

With $CK=BF$ as before, construct rectangle $CKND$ whose side $KN$ intersects $AG$ , $AD$ at $L$, $J$, respectively, forming$$\triangle AJL\sim\triangle NJD$$Since $ND=AF$, and $AF>AL$ (proof?), then in area$$\triangle NJD>\triangle AJL$$ and thus the triangle by which rectangle $CKND$ exceeds quadrilateral $ADCM$ is greater than the triangle by which $ADCM$ exceeds $CKND$.

And $CKND$ further exceeds $ADCM$ by concave quadrilateral $CKLM$.

Therefore, $CKND>ADCM$, and$$AB\cdot CD>2ADCM$$Note: As $AG$ increases, $AL$ and $ND$ converge toward equality, and area $CKLM$ converges to zero, so that in the limit$$AB\cdot CD=2ADCM$$

III. Returning to the first case, and keeping $BC$ fixed, if instead of increasing we diminish altitude $AG$, as in the next figure, since right triangles $AJD$, $NJL$ are similar, and $\triangle JNL>\triangle JAD$ (proof?), then $CKND$ exceeds $ADCM$ in area by$$\triangle JNL-\triangle JAD+CKLM$$and again$$AB\cdot CD>2ADCM$$ Note: Unlike the previous case, in which triangles $NJD$ and $AJL$ converge to equality as $AG$ increases, here$$\lim_{AG \to 0}{\frac{\triangle AJD}{\triangle NJL} \to 0}$$as both triangles vanish, while quadrilateral $MCKL$ approaches congruence with rectangle $CKND$.

Thus it seems$$\lim_{AG \to 0}{\frac{ADCM}{CKND} \to 0}$$ and the limiting value of $CKND$ is $$\lim_{AG \to 0}{\frac{AB}{2}\cdot {CD}}=\frac{CG^2}{2}$$

Conclusion: In general then, for isosceles $\triangle ABC$ $$AB\cdot CD\ge 2ADCM$$but it seems$$AB\cdot CD=2ADCM$$only at the limit in the second case, as $AG\to\infty$.

Answer 4

Let $A',B',C'$ be the mid points of the sides. Let $C''$ be the reflection of $C$ w.r.t. $C'$. (So $C''BCA$ is a parallelogram.)

The points so far are introduced in the order $A,B,C;A',B',C';M;C'',D$.

It is useful to change this "unnatural", order and moreover use only the following points in the following order: $$ C,C'';M,C';D;A\ . $$ Here $C,C''$ are arbitrary. Construct $C''$, the mid point of $CC'$, and $M$ on $CC'C''$ with $CM:CC''=1:3$. Consider now an arbitrary point $D$ on the perpendicular on $CMC'C''$ in $M$. Then finally $A$ is a free point on $C''D$.

We may want to remove $B$ from the picture, and use instead of $AB$ the value $2AC'$. See the first picture below for an orientation. We have some cases, that are best described by pictures, separated below by horizontal lines:

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$\bbox[beige]{\ (1)\ }$

In this case the point $D$ is between $A$ and $C''$. In particular, the projection of $A$ on the line $MC$ is on the half-line $[MC$, $C'$ is on the "other" half-line from $M$, and thus $AM<MC'$. We can conclude: $$ 2[ADMC]\le AM\cdot CD < AC'\cdot CD=\frac 12 AB\cdot CD\ . $$
Below we may and do assume that $A$ is on the half-line $DC''$.

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$\bbox[beige]{\ (2)\ }$

Let $C^\perp$ be the projection of $C'$ on $DAC''$. We consider now the case when $A$ is between $D$ and $C^\perp$.

Or if we remove all the unuseful stuff, and denote by $x,h$ the length of the segments $MC'$ and $MD$:

We can compute easily in the above picture $$ \begin{aligned} \frac{[C''C'C^\perp]}{[C''MD]} &=\left(\frac{C'C''}{C'D}\right)^2 =\frac{9x^2}{h^2 +16x^2}\ , \\ [C''C'C^\perp] &= \frac{9x^2}{h^2 +16x^2}\cdot 2xh\ ,\\ [C''MC^\perp] &= \frac 43[C''C'C^\perp]=\frac{12x^2}{h^2 +16x^2}\cdot 2xh\ ,\\ [C^\perp MCD] &= [C''CD] - [C''MC^\perp] = 3xh - \frac{8x^2}{h^2 +16x^2}\cdot 3xh = 3xh \cdot \frac{h^2 + 8x^2}{h^2 +16x^2} \ ,\\ C'C^\perp &= h\cdot \frac{3x}{\sqrt{h^2+16x^2}}\ ,\\ CD &= \sqrt{h^2 + 4x^2}\ ,\\ C'C^\perp \cdot CD &= 3xh\cdot \frac{\sqrt{h^2 + 4x^2}}{\sqrt{h^2+16x^2}}\ , \\[4mm] \left(\frac{2\; C'C^\perp\cdot CD}{2[C^\perp MCD]}\right)^2 & = \frac{h^2+4x^2}{h^2+16x^2}\cdot\frac{(h^2+16x^2)^2}{(h^2+8x^2)^2} = \frac{(h^2+4x^2)(h^2+16x^2)}{(h^2+8x^2)^2} \\ &>1\ .\qquad \text{ From here we obtain:} \\[4mm] 2\; C'C^\perp\cdot CD &>2[C^\perp MCD]\ ,\qquad\text{ and now finally:} \\[4mm] AB\cdot CD &= 2\; AC'\cdot CD\\ &>2\; C'C^\perp\cdot CD\\ &>2[C^\perp MCD]\\ &>2[A MCD]\ . \end{aligned} $$

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$\bbox[beige]{\ (3)\ }$

The case $A$ in the half-line $[C^\perp C''$ is the remained case:

Let us denote by $y$ the length of $AC^\perp$. Then the area $[AMCD]$ is the area of the triangle $[AMC^\perp]$ plus the already computed area $[C^\perp MCD]$. Note that $A$ may lie beyond $C''$, then $AMCD$ is no longer convex, but also in this case the relation is valid.

We then compute:

$$ \begin{aligned}{} [AMCD] &= [AMC^\perp] + [C^\perp MCD] =\frac y{C''D}{[C''MD]} + [C^\perp MCD] \\ &=\frac y{\sqrt{h^2 +16x^2}}\cdot 2xh + 3xh \cdot \frac{h^2 + 8x^2}{h^2 +16x^2} \\ &=\frac{xh}{h^2 +16 x^2}\left( \ 2y\sqrt{h^2+16x^2}+ 3(h^2+8x^2)\ \right)\ , \\[3mm] AB^2 &=4AC'{}^2 =4(y^2+C'C^\perp{}^2) =4\left(y^2 + h^2\cdot\frac {9x^2}{h^2+16x^2}\right)\ , \\ (AB\cdot CD)^2 &= 4\left( y^2 + h^2\cdot\frac {9x^2}{h^2+16x^2}\right)(h^2+4x^2)\ , \\[4mm] \left( \frac{AB\cdot CD}{2[AMCD]} \right)^2 &= \frac{(AB\cdot CD)^2}{4[AMCD]^2} \\ &= \frac {(16x^2y^2 + 9h^2x^2 + h^2y^2)(h^2+4x^2)\ /\ (h^2+16x^2)} {x^2h^2 \left( \ 2y\sqrt{h^2+16x^2}+ 3(h^2+8x^2)\ \right)^2\ /\ (h^2+16x^2)^2} \\ &= \frac {(16x^2y^2 + 9h^2x^2 + h^2y^2)\; (h^2+4x^2)\;(h^2+16x^2)} {x^2h^2 \left( \ 2y\sqrt{h^2+16x^2}+ 3(h^2+8x^2)\ \right)^2} \ . \end{aligned} $$ We want to show that the last expression is $\ge 1$. So consider the numerator of $\displaystyle \left( \frac{AB\cdot CD}{2[AMCD]} \right)^2 -1$ as a function of $y$ of degree two, explicitly $$ Ay^2 +By +C = (16x^2y^2 + 9h^2x^2 + h^2y^2)\; (h^2+4x^2)\;(h^2+16x^2)\\ \qquad\qquad\qquad - x^2h^2 \left( \ 2y\sqrt{h^2+16x^2}+ 3(h^2+8x^2)\ \right)^2\ . $$ Then we isolate $A,B,C$, and compute $$ \begin{aligned} A &= (h^2+4x^2)\;(h^2+16x^2)^2 - 4x^2h^2 (h^2+16x^2)^2 \\ &= (h^2+16x^2) (h^2+8x^2)^2 \ ,\\ B &= - x^2h^2 \cdot 4\sqrt{h^2+16x^2}\cdot 3(h^2+8x^2) \ ,\\ C&= 9x^2h^2\; (h^2+4x^2)\;(h^2+16x^2) - x^2h^2 \cdot 9(h^2+8x^2)^2 \\ &=36 x^4 h^4 \ ,\\ B^2 - 4AC &= 0\ , \end{aligned} $$ which means that $$ Ay^2+By+C= \Big(\ (h^2+8x^2)\sqrt{h^2+16x^2}\; y - 6x^2h^2\ \Big)^2 \ge 0 $$ makes indeed the needed quantity also non-negative.

$\square$

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Final notes: The value of $y$ leading to an equality is $$ y^* = \frac{6x^2h^2}{(h^2+8x^2)\sqrt{h^2+16x^2}}\ , $$ and i could not see a natural argument producing this quantity. In the expression there are of course as factors the lengths of segments like the obvious $h$ and some multiples of $x$, and the slightly less obvious
$C''D$. I could not see how to move using similarities the segment $CD$ from its direction to one on the line $DC''$.

It is late here in a hard day's night, so i have to submit here, possibly come back to this point if there is still something to say.

Regarding the way the solution was typed, yes, the first two cases could have been also compressed to fit inside the third and last case, but in conditions of an olympiad i would have lost a lot of points, so it is safe for the sake of didactics and clearness to draw more and type more...

Above, synthetic geometry was used as far as possible, and because i could not see any geometric sense for the ingredients of the inequality to be shown, i considered it as a geometric inequality. It was parametrized geometrically. We can solve it algebraically and possibly hide this solution if it give geometrical insight, but it did not. So the last algebraic step remained. In fact, initially i wanted to get one/any solution by using barycentric coordinated. (Because, as a speculation, an inequality of the same shape (may) should happen for all fixed constant barycentrics - the problem uses $M=[1:1:1]$. And because most of the solution can be recovered in this framework.) But the explicit wish of a synthetic approach was leading to the above. Still, this may be a special instance of a more general inequality.

Answer 5

Let $M$ be the centroid of triangle $ABC$. $D$ is a point on the line passing through $A$ and parallel to $BC$ such that $\,\angle CMD=90\unicode{176}.\,$ Let $\,S=[AMCD]\,.$
Prove that $\,AB\cdot CD\geqslant2S\,.$

Proof :

Without loss of generality, we can suppose that the coordinates of the vertices of the triangle $ABC$ are :
$A(x,y)\,,\,$ where $\,y>0\,,\;$ $B(0,0)\,,\,$ $C(1,0)\,.$

Consequently, the coordinates of the centroid $M$ are :
$M\left(\dfrac{x+1}3,\dfrac y3\right).$

The equation of the line $r$ passing through $M$ and perpendicular to the segment $CM$ is :

$r)\;\;\;Y=\dfrac{2-x}y\left(X-\dfrac{x+1}3\right)+\dfrac y3$

whereas the equation of the line $s$ passing through $A$ and parallel to $BC$ is the following one :

$s)\;\;\;Y=y\;.$

Since $\,r\bigcap s\equiv D\,,\,$ the lines $r$ and $s$ are not parallel, hence it results that $\;2-x\neq0\;\;\;\color{blue}{(1)}\;,\;$ moreover the coordinates of the point $D$ are :

$D\left(\dfrac{2+x+2y^2-x^2}{3(2-x)},y\right).$

The lengths of the segments $AB$ and $CD$ are :

$AB=\sqrt{x^2+y^2}\;\;,$

$\begin{align}CD&=\sqrt{\left[\dfrac{2y^2-(2-x)^2}{3(2-x)}\right]^2\!\!+y^2}=\\[3pt]&=\dfrac1{3|2-x|}\sqrt{\left[2y^2\!-(2-x)^2\right]^2\!+9y^2(2-x)^2}=\\[3pt]&=\dfrac1{3|2-x|}\sqrt{\left[2y^2\!+(2-x)^2\right]^2\!+y^2(2-x)^2}=\\[3pt]&=\dfrac1{3|2-x|}\sqrt{y^2(x-2)^2\!+\left[2y^2\!+(2-x)^2\right]^2}\;.\end{align}$

Let $x_A$ and $x_D$ be the abscissas of the points $A$ and $D$.

There are two possible cases :

$1)\;\;x_A\leqslant x_D\;\;,$

$2)\;\;x_A>x_D\;.$

In the first case, it results that

$\begin{align}2S&=2S_{AMC}+2S_{ACD}=\dfrac23S_{ABC}+y(x_D-x_A)=\\[3pt]&=\dfrac y3+y(x_D-x_A)=\dfrac y3\big[1+3(x_D-x_A)\big]=\\[3pt]&=\dfrac y3\left[1+\dfrac{2-5x+2x^2+2y^2}{2-x}\right]=\\[3pt]&=\dfrac{2y\left(2-3x+x^2+y^2\right)}{3(2-x)}\,.\end{align}$

By applying the Cauchy–Schwarz inequality, we get that

$\begin{align}AB\cdot CD&=\dfrac{\sqrt{x^2+y^2} \sqrt{y^2(x-2)^2\!+\left[2y^2\!+(2-x)^2\right]^2}}{3|2-x|}\geqslant\\[3pt]&\geqslant\dfrac{\bigg| x\big[y(x-2)\big]+y\!\left[2y^2\!+(2-x)^2\right]\bigg|}{3|2-x|}=\\[3pt]&=\left|\dfrac{2y\left(2-3x+x^2+y^2\right)}{3(2-x)}\right|\geqslant2S\;.\end{align}$

In the second case, it results that

$S=\begin{cases}S_{ADM}\!+\!S_{AMC}<S_{ADC}\!+\!S_{AMC}&\text{if }M\!\not\in\!\triangle ADC\\[3pt]<S_{ADC}<S_{ADC}\!+\!S_{AMC}&\text{if }M\!\in\!\triangle ADC\end{cases}$

Consequently ,

$\begin{align}2S&<2S_{ADC}+2S_{AMC}=y(x_A-x_D)+\dfrac23S_{ABC}=\\[3pt]&=y(x_A-x_D)+\dfrac y3=\dfrac y3\big[3(x_A-x_D)+1\big]=\\[3pt]&=\dfrac y3\left[\dfrac{5x-2-2x^2-2y^2}{2-x}+1\right]=\\[3pt]&=\dfrac{2y\left(2x-x^2-y^2\right)}{3(2-x)}=\dfrac{2y\big[x(2-x)-y^2\big]}{3(2-x)}\,.\end{align}$

Moreover ,

$\begin{align}AB\cdot CD&=\dfrac{\sqrt{x^2+y^2} \sqrt{y^2(x-2)^2\!+\left[2y^2\!+(2-x)^2\right]^2}}{3|2-x|}=\\[3pt]&=\dfrac{\sqrt{x^2\!+y^2}\sqrt{4y^4\!+5y^2(2-x)^2\!+(2-x)^4}}{3|2-x|}>\\[3pt]&>\dfrac{\sqrt{x^2\!+y^2}\sqrt{4y^4\!+4y^2(2-x)^2}}{3|2-x|}=\\[3pt]&=\dfrac{\sqrt{x^2\!+(-y)^2}\,\sqrt{4y^2(2-x)^2\!+\left(2y^2\right)^2}}{3|2-x|}\!\!\!\!\!\!\!\!\!\!\!\!\underset{\overbrace{\text{ Cauchy–Schwarz }\\\text{ inequality }}}{\geqslant}\\[3pt]&\geqslant\dfrac{\big|2xy(2-x)-2y^3\big|}{3|2-x|}=\dfrac{\bigg|2y\big[x(2-x)-y^2\big]\bigg|}{3|2-x|}=\\[3pt]&=\left|\dfrac{2y\big[x(2-x)-y^2\big]}{3(2-x)}\right|\geqslant\dfrac{2y\big[x(2-x)-y^2\big]}{3(2-x)}>2S\;.\end{align}$

Hence, in any case we have proved that $\,AB\cdot CD\geqslant2S\,.\\[10pt]$

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Addendum :

Now, we will prove that

if the coordinates of the vertices of the triangle $ABC$ are $\,A(x,y)\,,\,$ where $\,y>0\,,\;$ $B(0,0)\,,\,$ $C(1,0)\,,\;$ then

$AB\cdot CD=2S\iff\begin{cases}x^3+(x+2)y^2-4x^2+4x=0\\[3pt]-2<x<0\end{cases}$

Proof :

$AB\cdot CD=2S\;$ implies $\;x_A\leqslant x_D\,,\;$ otherwise it would result $\,AB\cdot CD>2S\;$ (see the second case of the previous proof).

Let $\;a=x\,,\,$ $b=y\,,\,$ $c=y(x-2)\,,\,$ $d=2y^2\!+(2-x)^2\,.$

It results that

$ac+bd=2y\left(2-3x+x^2+y^2\right)\;\;,$

$ad-bc=x^3+(x+2)y^2-4x^2+4x\;\;.$

Moreover, according to the first case of the previous proof,
$AB=\sqrt{a^2+b^2}\;,\;$ $CD=\dfrac{\sqrt{c^2+d^2}}{3|2-x|}\;,\;$ $2S=\dfrac{ab+cd}{3(2-x)}\;.$

From the equalities

$\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\;\;,$

$AB^2\!\cdot CD^2=4S^2\;\;,$

it follows that

$ad-bc=0\;\;,\;\;$ that is

$x^3+(x+2)y^2-4x^2+4x=0\;\;,\;\;$ which implies

$-2<x<0\;.$

On the other hand

$\begin{cases}x^3+(x+2)y^2-4x^2+4x=0\\[3pt]-2<x<0\end{cases}\quad$ implies

$x_A-x_D=\dfrac{5x-2-2x^2-2y^2}{3(2-x)}<0\;\;,\;\;$ that is ,

$x_A<x_D\quad$ ( first case of the previous proof ).

Let $\;a=x\,,\,$ $b=y\,,\,$ $c=y(x-2)\,,\,$ $d=2y^2\!+(2-x)^2\,.$

According to the first case of the previous proof,
$AB=\sqrt{a^2+b^2}\;,\;$ $CD=\dfrac{\sqrt{c^2+d^2}}{3|2-x|}\;,\;$ $2S=\dfrac{ab+cd}{3(2-x)}\;.$
Actually , $\;CD=\dfrac{\sqrt{c^2+d^2}}{3(2-x)}\;$ because $\;-2<x<0\;$ implies $\;2-x>0\;.$

Since

$ac+bd=2y\left(2-3x+x^2+y^2\right)>0\;\;,$

$ad-bc=x^3+(x+2)y^2-4x^2+4x=0\;\;$ and

$\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\;\;,$

it follows that

$\sqrt{a^2+b^2}\sqrt{c^2+d^2}=ab+cd\;\;,\;\;$ hence ,

$AB\cdot CD=2S\,.$

Answer 6

Using vectors

Calling

$$ A=(x_a,y_a), B=(0,0), C=(x_c,0), D = (x_d,y_a) $$

we have

$$ M = \frac 13(A+B+C)\\ S = \frac 12\|\frac{4}{3} y_a \left(x_a -\frac{1}{3} (x_a+x_c)\right)+\frac{1}{3} y_a \left(\frac{x_a+x_c}{3}-x_c\right)+2 y_a(x_d-x_a)+y_a (x_c-x_d)\| $$

now from $\|D-M\|^2+\|C-M\|^2=\|D-C\|^2$ we obtain

$$ y_a^2=\frac 12\left(x_a-2x_c\right)\left(x_a+x_c-3x_d\right)\ \ \ \ \ \ (*) $$

and from the condition

$$ \|A-B\|^2\|C-D\|^2\ge 4 S^2 \Rightarrow(x_a^2+y_a^2)\left(y_a^2+(x_c-x_d)^2\right)\ge \frac 19(x_c+3(x_d-x_a))^2y_a^2 $$

Using $(*)$ we got

$$ \left(3x_a^2+2x_c(x_c-3x_d)-x_a(x_c+3x_d)\right)^2\ge 0 $$

meaning that the statement it is true.