<AEF=<CAD if AB//CD, BC//AD, AE=AD - 2021 Rusanovsky Lyceum Olympiad 7.6

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High School Math Grades 9-12, Ages 13-18

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#1 h Jul 8, 2021, 7:29 AM

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It is known that in the quadrilateral $ABCD$ (AB>AD), $AB \parallel CD$ , $BC\parallel AD$ . Point $E$ lies on the side $AB$ such that $AE=AD$ . In the extension of DA beyond point A, we take a point F such that $BF \perp DE$ . Prove that the angles $AEF$ and $CAD$ are equal.

(M. Plotnikov)

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#2 h Jul 9, 2021, 7:36 PM

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Given $A(0,0),B(b,0),D(d\cos \alpha,d\sin \alpha),C(b+d\cos \alpha,d\sin \alpha),E(d,0)$ .

Slope of the line $DE\ :\ m_{DE}=\frac{\sin \alpha}{\cos \alpha-1}$ ,
equation of the line $BF\ :\ y=\frac{1-\cos \alpha}{\sin \alpha}(x-b)$ .
This line cuts the line $AD\ :\ y=\tan \alpha \cdot x$ in the point $F(-b\cos \alpha,-b\sin \alpha)$ .

Slope of the line $EF\ :\ m_{EF}=\frac{b\sin \alpha}{b\cos \alpha+d} \quad(1)$ .

Slope of the line $AC\ :\ m_{AC}=\frac{d\sin \alpha}{b+d\cos \alpha}$ , slope of the line $AD\ :\ m_{AD}=\tan \alpha$ ,
then $\tan \angle CAD =\frac{m_{AD}-m_{AC}}{1+m_{AD} \cdot m_{AC}}=\frac{b\sin \alpha}{b\cos \alpha+d}$ , see (1), so $\angle CAD=\angle AEF$ .

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