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For reference: In the figure, ABCD is a parallelogram and $ \measuredangle ABE = 100^o$ , $AE = AD$ and $FC = CD$. Calculate x.

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My progress:

In this one I couldn't develop much because $FC$ and $EB$ are not parallel...I think the "output" is by some auxiliary construction

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by geogebra

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    $\begingroup$ You say that $\angle ABE=145$, but your diagram suggests that it is $100$. Is there a typo? $\endgroup$ Aug 20, 2021 at 14:33
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    $\begingroup$ $\Delta EAB \cong \Delta BCF$ by SAS congruency criterion $\endgroup$ Aug 20, 2021 at 14:50
  • $\begingroup$ @peta arantes, in figure $\angle ABE=100 ^o$ which one is correct? $\endgroup$
    – sirous
    Aug 20, 2021 at 16:35
  • $\begingroup$ @AlanAbraham..my mistake, thanks for the alert..I've corrected the statement $\endgroup$ Aug 20, 2021 at 18:52
  • $\begingroup$ @sirous my mistake, thanks for the alert..I've corrected the statement $\endgroup$ Aug 20, 2021 at 18:52

3 Answers 3

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In $\Delta EAB$ and $\Delta BCF$ $\begin{align} EA&=BC &(\because AD=BC, \text{ opposite sides of parallelogram})\\ AB&=CF & (\because CD=AB, \text{ opposite sides of parallelogram)}\\ \angle EAB&=\angle BCF=y \text{ (let)} & (\because 90°-\angle BAC=90°-\angle BCD)\\ \therefore \Delta EAB &\cong \Delta BCF &\text{ (by SAS congruency criterion)}\\ \implies \angle EBA&=\angle BFC= 100°&\text{ (corresponding angles of congruent triangles)}\\ \text{and}\qquad EB&=BF &\text{ (corresponding sides of congruent triangles)}\\ \end{align}$

In parallelogram $ABCD$,

$\begin{align}&\angle BAD =90°-y\\ \implies &\angle ABC=180°-(90°-y)=90°+y \end{align}$

In triangle $BCF$, $\angle FBC= 180°-(100°+y)=80°-y$

In triangle $EBF$, $\angle BEF=\angle BFE=x-100° \; (\because EB=BF)$

$\implies \angle EBF=180°-2(x-100°)=380°-2x$

Sum of all angles around point $B$ $= 100°+(90°+y)+(80°-y)+(380°-2x)=360°$

$\therefore x=145°$

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  • $\begingroup$ $\angle EBA= 145°$ is not possible $\endgroup$ Aug 20, 2021 at 15:19
  • $\begingroup$ @Aman..great resolution.. grateful $\endgroup$ Aug 20, 2021 at 18:54
  • $\begingroup$ @petaarantes you are welcome $\endgroup$ Aug 21, 2021 at 1:44
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enter image description here

All you need is to show points F, B, D and C are cyclic, then we have:

$\angle ADB=45$

$\angle FBD=90$

$\rightarrow \angle FAB=35\rightarrow \angle BAD=\angle BCD=55=\angle BFD$

$\Rightarrow \angle ABD=80$

$\angle BFC=55+45=100$

$\angle BEF=45$

$\angle FBE=360-(100+80+90)=90$

$\Rightarrow \angle EFB=45$

Finally:

$x=45+100=145$

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  • $\begingroup$ @sirous...EB is not parallel to FC so I don't think it's cyclic..see my last figute..$\rightarrow \angle FAB=35$ I don't understand $\endgroup$ Aug 20, 2021 at 19:17
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Another solution.

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Perform a $90^{\circ}$ clock-wise rotation around $D$ and let the rotated image of $A$ be denoted by $H$. Then $$AD = DH = AE$$ which means that $ADHE$ is a square. If you rotate $90^{\circ}$ counter-clock-wise around $D$ the segment $CD$, you get the segment $GD$ such that $GD = CD$ and $GD \, \perp \, CD$. Hence, $CDGF$ is a square. Because of these new extra constructions, during the $90^{\circ}$ clock-wise rotation around $D$, The points $A$ and $G$ are rotated to the points $H$ and $C$ and consequently, the triangle $ADG$ is rotated to the triangle $HDC$ which means that $$\angle\, HCD = \angle\, AGD$$ $$HC = AG$$ $$HC \, \perp \, AG$$ Now, if you translate triangle $HDC$ along $DA$ you get as a parallel translated image the triangle $EAB$. Furthermore, if you translate triangle $ADG$ along $AB$ you get as a parallel translated image the triangle $BCF$. Therefore, $$\angle\, EBA = \angle \, BFC = 100^{\circ}$$ $$EB = FB$$ $$EB \, \perp \, FB$$ Hence, triangle $EBF$ is right-angled and isosceles, which means that $$\angle \, EFB = 45^{\circ}$$ Now putting all these angles together $$x = \angle \, EFC = \angle\, EFB + \angle \, BFC = 45^{\circ} + 100^{\circ} = 145^{\circ}$$

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