Congurence of Triangles: Problems with Solutions

Solution:
In triangle $\Delta BCA$ we know that $\measuredangle CAB=180^{o}-140^{o}-10^{o}=30^{o}$

While in triangle $\Delta EFD$ we know that $\measuredangle DFE=180^{o}-140^{o}-30^{o}=10^{o}$

Now, we can set the congruence $\Delta BCA\cong \Delta EFD$ using the ASA postulate, provided that

$\left\{ \begin{array}{c} \measuredangle ACB\cong \measuredangle DEF \\ AC\cong DF \\ \measuredangle CAB\cong \measuredangle FDE \end{array} \right\} \textbf{ASA} \Longrightarrow \Delta BCA\cong \Delta EFD$

Solution:
Given that $ABCD$ is a parallelogram, point $E$ is the midpoint of each diagonal. According to this, we can apply the SAS congruence postulate in triangles $\Delta DEA$ and $\Delta BEC$ to conclude that $\Delta DEA\cong \Delta BEC$ provided that:

$\left\{ \begin{array}{c} DE\cong BE \\ \measuredangle AED\cong \measuredangle CEF \\ AE\cong CE \end{array} \right\} \textbf{SAS} \Longrightarrow \Delta DEA\cong \Delta BEC$

Solution:
According to the given data and using the SAS postulate, we can say these two triangles are congruent $\Delta ADE\cong \Delta BCD$ because the following conditions are met:

$\left\{ \begin{array}{c} BC\cong AD \\ \measuredangle ADE\cong \measuredangle DCB=\alpha \\ CD\cong DE \end{array} \right\} \textbf{SAS} \Longrightarrow ADE\cong \Delta BCD$

This congruence allows us to state that $\measuredangle EAD\cong \measuredangle CBD=20^{o}$

Now, we can find the angle $\alpha$. Let us see the triangle $\Delta ADE$ where $115^{o}+20^{o}+\alpha =180^{o}$ from this equation we can find that $\alpha =45^{o}$

Solution:
In the figure, [tex]\angle B=\angle E=40^\circ[/tex] and [tex]\angle A\cong\angle D[/tex], so two pairs of corresponding angles are already congruent. To apply the ASA postulate, we also need the side between these angles: [tex]AB[/tex] in [tex]\triangle ABC[/tex] and [tex]DE[/tex] in [tex]\triangle DEF[/tex]. If [tex]AB\cong DE[/tex], then
[tex]\angle A\cong\angle D,\quad AB\cong DE,\quad \angle B\cong\angle E\ \Rightarrow\ \triangle ABC\cong\triangle DEF[/tex]
The other options do not help: [tex]\angle B\cong\angle E[/tex] is already known; [tex]\angle C\cong\angle F[/tex] follows from the angle sum [tex]180^\circ[/tex], and three pairs of congruent angles do not guarantee congruent triangles (the triangles may have different sizes); [tex]AC\parallel DF[/tex] says nothing about the lengths of the sides.

Solution:
In the figure, [tex]\angle B=\angle E=40^\circ[/tex] and [tex]\angle A\cong\angle D[/tex]; moreover, [tex]AB\cong DE[/tex]. The side [tex]AB[/tex] lies between the angles [tex]A[/tex] and [tex]B[/tex], and the side [tex]DE[/tex] lies between the angles [tex]D[/tex] and [tex]E[/tex], so these are the included sides:
[tex]\angle A\cong\angle D,\quad AB\cong DE,\quad \angle B\cong\angle E\ \Rightarrow\ \triangle ABC\cong\triangle DEF[/tex]
Two angles and the included side are congruent, so the triangles are congruent by the ASA postulate (angle-side-angle). The other criteria do not apply: SAS, SSS and HL (hypotenuse-leg) all need at least two pairs of congruent sides (HL also needs right triangles), but here only one pair of sides is known to be congruent.

Solution:
Provided the congruence $\Delta ABC\cong \Delta DBE$, the following congruences are met:

$\measuredangle CAB\cong \measuredangle EDB=36^{0}$
$\measuredangle CBA\cong \measuredangle EBD$

Now, we also know that: $\measuredangle CBA+\measuredangle CBE+\measuredangle EBD=180^{o}$

So, $2\measuredangle EBD=180^{o}-\measuredangle CBE=180^{o}-20^{o}=160^{o}$

This way, $\measuredangle EBD=80^{o}$ and then in triangle $\Delta DBE$ we have that:

$\measuredangle DEB+\measuredangle EBD+\measuredangle EDB=180^{o}$. Finally, we can find the angle measure:

$\measuredangle DEB=180^{o}-\measuredangle EBD-\measuredangle EDB=180^{o}-80^{o}-36^{o}=64^{o}$

$\measuredangle DEB=64^{o}$