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The triangle $ABC$ is isosceles at $A$ with $\angle BAC=20^o$. The points $D$ and $E$ are on the sides $[AB]$ and $[AC]$, respectively. We know that $\angle EBC=40^o$ and $\angle BCD=50^o$. What is the measure of $\angle BED$?

What I did: since $ABC$ is isosceles at $A$ and $\angle BAC=20^o$, we get that $\angle ABC=\angle BCA=80^o$. Therefore, we have $\angle DBE=40^o$ and $\angle BCD=50^o$. Moreover, the lines $(EB)$ and $(CD)$ are perpendicular. It follows that $\angle BEC=60^o$ and $\angle BDC=50^o$.

All the angles of the "butterfly" BDCE are known, and we know the angle $\angle EBC$. So the $\angle BED$ should not be too complicated to get, but I am stuck. Any idea?

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  • $\begingroup$ Do you want a geometric solution or is a trigonometric solution fine? $\endgroup$ Commented Mar 21, 2021 at 4:29
  • $\begingroup$ @dodoturkoz: good remark. I would probably accept an answer using the sum of angles in a triangle/quadrangle. But I will certainly upvote an answer using trigonometry. $\endgroup$ Commented Mar 21, 2021 at 4:32
  • $\begingroup$ This is a version of the Langley's Adventitious Angles problem. You can solve via trigonometry using Trigonometric Ceva's Theorem. $\endgroup$ Commented Mar 21, 2021 at 4:43

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As $\angle BCD = 50^0, \angle CBE = 40^0$, we have $\angle BFC = 90^0$.

Please note as $\angle DBF = \angle CBF$ and $BF \perp CD$, $BF$ is perpendicular bisector of $CD$. So, $DF = FC$ and $EF \perp CD$.

That leads to $\triangle DFE \cong \triangle CFE$. So, $\angle CDE = \angle DCE = 30^0$ and $\angle BED = \angle BEC = 60^0$.

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This is an easier version of the classical problem since one of the lines is an angle bisector and can be solved using congruences.

Let $BE$, $CD$ intersect in $O$. $\triangle BOD \cong \triangle BOC$ using $ASA$ since $BE$ is angle bisector. Hence $OD=OC$.

Thus $\triangle DOE \cong \triangle COE$ using $SAS$. Hence $$\angle BED = \angle BEC=60^{\circ}$$

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    $\begingroup$ yesterday's high school geometry solution was nice :) $\endgroup$ Commented Mar 21, 2021 at 5:19
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    $\begingroup$ Thanks. It was a nice problem :) $\endgroup$ Commented Mar 21, 2021 at 5:20

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