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Practice
Congurence of Triangles
Congurence of Triangles: Problems with Solutions
Congurence of Triangles - SAS, SSS, ASA, SSA
Problem 1
According to the information shown in the graph, which of the following assertions are true?
I) $\measuredangle ACB\cong \measuredangle DEF$
II) $AB\cong EF$
III) $\Delta BCA\cong \Delta EFD$
Only I
Only II
Only III
Only I and III
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
$\left\{ \begin{array}{c} \measuredangle ACB\cong \measuredangle DEF \\ AC\cong DF \\ \measuredangle CAB\cong \measuredangle FDE \end{array} \right\} \textbf{ASA} \Longrightarrow \Delta BCA\cong \Delta EFD$
Problem 2
In the following figure, $AD||BC$ and $DC||AB$. Which of the next congruences are always true?
I) $\Delta DEA\cong \Delta BEC$
II) $\Delta DEC\cong \Delta DEA$
III) $\Delta DBC\cong \Delta CAB$
Only I
Only II
Only III
Only I and II
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:
$\left\{ \begin{array}{c} DE\cong BE \\ \measuredangle AED\cong \measuredangle CEF \\ AE\cong CE \end{array} \right\} \textbf{SAS} \Longrightarrow \Delta DEA\cong \Delta BEC$
Problem 3
In triangle $\Delta ABC$ from the following figure, $\measuredangle CBD=20^{o}$. $BC\cong AD$ and $CD\cong DE$, what's the angle measure $\measuredangle \alpha $ ?
$20^{o}$
$45^{o}$
$60^{o}$
$30^{o}$
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 $\alpha =45^{o}$
Problem 4
What's the data missing so that we can state that both triangles in the following figure are congruent?
What congruence postulate is possible to use to state that $\Delta ABC\cong \Delta DEF$ ?
$\measuredangle B\cong \measuredangle E$
$\measuredangle C\cong \measuredangle F$
$AC||DF$
$AB\cong DE$
Solution:
Provided that $\measuredangle B\cong \measuredangle E=40^{o}$ , through the
ASA
postulate
we can say that: $\left\{ \begin{array}{c} \measuredangle ABC\cong \measuredangle DEF \\ AB\cong DE \\ \measuredangle CAB\cong \measuredangle FDE \end{array} \right\} ASA \Longrightarrow \Delta ABC\cong \Delta DEF$
Problem 5
In the following figure, points $A,B,D$ are collinear(lie in the same straight line), $\Delta ABC\cong \Delta DBE,\measuredangle \alpha =36^{o}$ and $\measuredangle CBE=20^{o}$. What is the angle measure $\measuredangle DEB$ ?
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:
$\measuredangle DEB=180^{o}-\measuredangle EBD-\measuredangle EDB=180^{o}-80^{o}-36^{o}=64^{o}$
$\measuredangle DEB=64^{o}$
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