2
$\begingroup$

enter image description here

$D$ lies on the smaller of arc $AB$ of the circumcircle of isoceles triangle $ABC$ at $A$. $E$ is a point on line segment $BC$ such that $AD \perp DE$. The perpendicular bisector $DE$ cuts $AC$ at point $F$. Prove that $EF \parallel AB$.

At first, I thought I needed to let $G$ be the intersection of $AE$ and the perpendicular bisector of $DE$ so $\triangle ADG$ is an isosceles triangle at $G \implies \widehat{EAD} = \widehat{GDA}$. (Because $\triangle ADE$ is a right-angled at $D$.)

Secondly, I assumed letting $H = AB \cap DF$ and proved that $H$ lies on the perpendicular bisector of $AD \implies \widehat{BAD} = \widehat{FDA}$. (But I couldn't.)

Therefore, $\widehat{EAD} - \widehat{BAD} = \widehat{GDA} - \widehat{FDA} \implies \widehat{EAB} = \widehat{GDF} \implies \widehat{EAB} = \widehat{AEF}$ (Because $G$ lies on the perpendicular of $DE$.)

$\endgroup$
1
$\begingroup$

enter image description here Let the perpendicular bisector of $DE$ cuts $AB$ and $BD$ respectively at $G$ and $H$.

We have that $AD \parallel FH \implies \widehat{CAD} = \widehat{CFH}$.

$ACBD$ is a cyclic quadrilateral $\implies \widehat{CAD} + \widehat{CBD} = 180^\circ$.

That means $\widehat{CFH} + \widehat{CBH} = 180^\circ \implies FCBH$ is a cyclic quadrilateral $\implies \widehat{FCB} = \widehat{DHF}$.

But $FH$ is the perpendicular bisector of $DE \implies \widehat{DHF} = \widehat{EHF}$.

$\triangle ABC$ is an isosceles triangle at $A \implies \widehat{ABC} = \widehat{ACB}$.

That means $\widehat{GBE} = \widehat{EHG} \implies GEBH$ is a cyclic quadrilateral $\implies \widehat{EBH} = \widehat{FGE}$.

But $FG$ is the perpendicular bisector of $ED \implies \widehat{FGE} = \widehat{FGD}$.

$ACBD$ is a cyclic quadrilateral $\implies \widehat{CAD} + \widehat{CBD} = 180^\circ$.

That means $\widehat{FAD} + \widehat{FGD} = 180^\circ \implies AFGD$ is a cyclic quadrilateral $\implies \widehat{DAG} = \widehat{DFG}$.

But $FG$ is the perpendicular bisector of $DE \implies \widehat{DFG} = \widehat{EFG}$.

But $AD \parallel FH \implies AB \parallel FE$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.