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Source: 47-th Belarusian Mathematical Olympiad 1997

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#1 h May 5, 2017, 2:14 PM • 1 Y

Y by Adventure10

$Problem 2:$ Points $D$ and $E$ are taken on side $CB$ of triangle $ABC$ , with $D$ between $C$ and $E$ ,
such that $\angle BAE =\angle CAD$ . If $AC < AB$ , prove that $AC.AE < AB.AD$ .

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#2 h May 5, 2017, 2:41 PM • 2 Y

Y by Adventure10, Mango247

Condition to be proved is equivalent with $[sin(B)/sin(C)][sin((x+B)/sin(x+C)]$ where $B=ABC$ and $C=ACB, x=BAE$ .
$AC<AB$ is equivalent to $sin(B)/sin(C)<1$ and since $D$ is in middle of $E$ and $C$ we have that $2x+B+C<A+B+C=180$
same as $180-(x+C)>x+B$ => $sin((x+B)/sin(x+C)<1$ .

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#3 h Jun 2, 2017, 11:51 AM • 2 Y

Y by Adventure10, Mango247

Since, $AC<AB$ , $\angle B < \angle C$
Now, $\angle AED = \angle B + \angle BAE$
$< \angle C + \angle CAD$
$= \angle ADE$
Therefore, $AD<AE$
or, $AD.AC<AB.AE$ [since $AC<AB$ ]
or, $\frac{1}{2}$ . $AD.AC$ $sin(X)$ < $\frac{1}{2}$ . $AB.AE$ $sin(X)$ [where, $\angle X =\angle BAE= \angle CAD$ ]
or, $area ACD< area ABE$
or, $area ACE< area ABD$
or, $\frac{1}{2}$ . $AE.AC$ $sin(Y)$ < $\frac{1}{2}$ . $AD.AB$ $sin(Y)$ [where $\angle Y=\angle X + \angle DAE ] \implies AC.AE < AB.AD$

Done!

This post has been edited 8 times. Last edited by mugdhosnigdho, Jun 2, 2017, 12:30 PM
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