Spanish problem [antiparallels yield BE / CD = (c/b)^2]

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Spanish problem [antiparallels yield BE / CD = (c/b)^2]

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Source: Spanish Olympiad

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#1 h Feb 12, 2005, 10:33 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle. Points $D$ and $E$ are taken on the line $BC$ such that $AD$ and $AE$ are parallel to the respective tangents to the circumcircle at $C$ and $B$ . Prove that
$\frac{BE}{CD}=\left(\frac{AB}{AC}\right)^2$

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#2 h Feb 12, 2005, 12:09 PM • 2 Y

Y by Adventure10, Mango247

The problem is quite straightforward so I just point out main steps of solution:

1) Angle chasing [making use of those paralells and everything else you can find to get all the angles within the triangle

]

2) Sine law applied to $ABC$ , $AEB$ and $ACD$ - multiplying proper relations we get what we wanted

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#3 h Feb 15, 2005, 5:43 PM • 2 Y

Y by Adventure10, Mango247

warut_suk wrote:

$ABC$ is a triangle. $D$ , $E$ are points on the line $BC$ such that $AD$ , $AE$ are parallel to the tangents to the circumcircle at $C$ , $B$ . Show that $\frac{BE}{CD} = \left(\frac{AB}{AC}\right)^2$ .

The problem is almost trivial:

We will use directed angles modulo 180°.

Let y be the tangent to the circumcircle of triangle ABC at the point B. Then, by the tangent-chordal angle theorem, we have < (y; BC) = < BAC. On the other hand, by the conditions of the problem, the line AE is parallel to this tangent y to the circumcircle of triangle ABC at the point B; thus, < (AE; BC) = < (y; BC). Hence, < (AE; BC) = < BAC. In other words, < AEB = - < CAB. Also, trivially, < ABE = - < CBA. Thus, the triangles AEB and CAB are inversely similar; consequently, $\frac{BE}{AB}=\frac{BA}{CB}$ , so that $BE=AB\cdot\frac{BA}{CB}=AB\cdot\frac{AB}{BC}$ . Similarly, $CD=AC\cdot\frac{AC}{BC}$ . Thus,

$\frac{BE}{CD}=\frac{AB\cdot\frac{AB}{BC}}{AC\cdot\frac{AC}{BC}}=\frac{AB^2}{AC^2}=\left(\frac{AB}{AC}\right)^2$ ,

and the problem is solved.

Darij

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