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Let $ABCD$ be a right angle trapezoid, with angle $\hat{A} =90$ degrees and $\overline{AB}$ is parallel to $\overline{CD}$. Let $O$ be the intersection point of the diagonals $\overline{AC}$ and $\overline{BD}$. The parallel through $O$ to the basis intersects $\overline{AD}$ at $E$ (which lies on $\overline{AD}$). Prove that $\overline{EO}$ is the angle bisector of angle $\widehat{BEC}$. (I believe the proof must be answered using Side Angle Side similarity between triangle $CDE$ and triangle $BAE$)

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    $\begingroup$ Mindlessly taking the first few words in the problem statement is NOT a way to produce a question title that will tell anyone anything about what the question is about! $\endgroup$ – hmakholm left over Monica Mar 3 '14 at 16:06
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As $AB // EO // CD$, $\hat{CEO} = \hat{ECD}$ and $\hat{BEO} = \hat{EBA}$, so it's necessary to prove that $\hat{EBA} = \hat{ECD}$...

Let $\hat{EBA} = \alpha$, $\hat{ECD} = \beta$ and $F$ the intersection between $EO$ and $CB$. Notice that $ABO\cong CDO$ ($AB // CD$ and $\hat{AOB} = \hat{COD}$), so $$AB:CD=AO:OC=BO:OD$$ By applying $Thales'$ $Theorem$ you obtain that $$AE:ED=AO:OC$$ $$\to AO:OC=AB:CD=AE:ED$$ $$\to ED=\frac{AE\cdot CD}{AB}$$ Now, $$\tan\beta = \frac{ED}{CD} = \frac{AE\cdot CD}{AB}\frac{1}{CD} = \frac{AE}{AB} = \tan\alpha$$ $$\to \beta = \alpha$$ It can't be $\to \beta = \alpha \pm \pi$ because both $\beta$ and $\alpha$ are $\lt \frac{\pi}{2}$

$Q.E.D.$

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Hint: Prove that $$\frac{AE}{DE}=\frac{BO}{DO}=\frac{AB}{CD}.$$

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  • $\begingroup$ but $AE/ED=BO/DO$ so if this equals $BO/EO$ then $EO=DO$ which is not true $\endgroup$ – user126154 Mar 3 '14 at 16:15
  • $\begingroup$ @user126154 Sorry, I meant $BO/DO$ rather than $BO/EO$. You get the idea. I corrected the error. $\endgroup$ – S.B. Mar 3 '14 at 16:34
  • $\begingroup$ yes, sure. thanks $\endgroup$ – user126154 Mar 3 '14 at 16:41

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