Equivalence between sides - Portuguese MO, Problem 2, 2008

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Equivalence between sides - Portuguese MO, Problem 2, 2008

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Joao Pedro Santos

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Joao Pedro Santos

#1 h Aug 31, 2010, 5:21 PM • 2 Y

Y by Adventure10, Mango247

Let $AEBC$ be a cyclic quadrilateral. Let $D$ be a point on the ray $AE$ which is outside the circumscribed circumference of $AEBC$ . Suppose that $\angle CAB=\angle BAE$ . Prove that $AB=BD$ if and only if $DE=AC$ .

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#2 h Aug 31, 2010, 5:38 PM • 2 Y

Y by Adventure10, Mango247

$AEBC$ is cyclic and $\angle BAC=\angle BAE$ so we conclude that $CB=EB$ if $DE=AC$ then taking into acount that $\angle ACB=\angle BED$ we get that $\triangle ACB=\triangle BED$ so $AB=BD$ .If $BD=AB$ then $\triangle ABD$ is isosceles and one more time $\triangle ACB=\triangle BED$ (because $CB=BE$ $\angle ACB=\angle BED$ and $\angle DAB=\angle CAB=\angle EDB$ ).
So we are done.

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#3 h Aug 25, 2013, 1:52 PM • 1 Y

Y by Adventure10

We easily get $BEC$ is isosceles triangle with $BE=BC$ .Now from triangles $ABC$ and $BED$ we get $BE=BC$ , ${\angle}DEB={\angle}BCA$ ,now if $DE=AC$ then $AB=BD$ ,Hence we get the result.

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#4 h Jun 6, 2023, 1:09 AM

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Joao Pedro Santos wrote:

Let $AEBC$ be a cyclic quadrilateral. Let $D$ be a point on the ray $AE$ which is outside the circumscribed circumference of $AEBC$ . Suppose that $\angle CAB=\angle BAE$ . Prove that $AB=BD$ if and only if $DE=AC$ .

In both cases $\triangle BDE \cong \triangle BAC$

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