2008 JBMO Shortlist G6

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Source: 2008 JBMO Shortlist G6

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#1 h Oct 10, 2017, 4:40 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle with $\angle A<{{90}^{o}}$ . Outside of a triangle we consider isosceles triangles $ABE$ and $ACZ$ with bases $AB$ and $AC$ , respectively. If the midpoint $D$ of the side $BC$ is such that $DE \perp DZ$ and $EZ = 2 \cdot ED$ , prove that $\angle AEB = 2 \cdot \angle AZC$ .

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#2 h Feb 9, 2025, 5:08 PM

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Since $D$ is the midpoint of the side $BC$ , in the extension of the line segment $ZD$ , we take a point $H$ such that $ZD = DH$ . Then the quadrilateral $BHCZ$ is a parallelogram, and therefore we have
$BH = ZC = ZA. \tag{1}$
Also, from the isosceles triangle $ABE$ , we get
$BE = AE. \tag{2}$
Since $DE \perp DZ$ , $ED$ is the altitude and median of the triangle $EZH$ , and so this triangle is isosceles with
$EH = EZ. \tag{3}$
From (1), (2), and (3), we conclude that the triangles $BEH$ and $AEZ$ are equal. Therefore, they also have
$\angle BEH = \angle AEZ, \quad \angle EBH = \angle EAZ, \quad \text{and} \quad \angle EHB = \angle AZE. \tag{4}$
Putting $\angle EBA = \angle EAB = \omega$ , and $\angle ZAC = \angle ZCA = \varphi$ , then we have $\angle CBH = \angle BCZ = \text{C} + \varphi$ , and therefore, from the equality $\angle EBH = \angle EAZ$ , we get:
$360^\circ - \angle EBA - \text{B} - \angle CBH = \angle EAB + \text{A} + \angle ZAC$ $\Rightarrow 360^\circ - \text{B} - \omega - \varphi - \text{C} = \omega + \text{A} + \varphi$ $\Rightarrow 2(\omega + \varphi) = 360^\circ - \left(\text{A} + \text{B} + \text{C}\right)$ $\Rightarrow \omega + \varphi = 90^\circ$ $\frac{180^\circ - \angle AEB}{2} + \frac{180^\circ - \angle AZC}{2} = 90^\circ$
\]
$\Rightarrow \angle AEB + \angle AZC = 180^\circ. \tag{5}$
From the supposition $EZ = 2 \cdot ED$ , we get that the right triangle $ZEH$ has $\angle EZD = 30^\circ$ and $\angle ZED = 60^\circ$ . Thus, we have $\angle ZEH = 120^\circ$ .

However, since we have proved that $\angle BEH = \angle AEZ$ , we get that
$\angle AEB = \angle AEZ + \angle ZEB = \angle ZEB + \angle BEH = \angle ZEH = 120^\circ. \tag{6}$
From (5) and (6), we obtain that $\angle AZC = 60^\circ$ , and thus
$\angle AEB = 2 \cdot \angle AZC. \tag{7}$

This post has been edited 2 times. Last edited by mathbetter, Feb 9, 2025, 5:15 PM

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