How to find if of acute / obtuse isosceles triangles are congruent - Intermediate Geometry

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Question

Refer to the above diagram. $\overline{AD} \cong \overline{BC}$ and $\overline{AC} \cong \overline{BD}$ . By what statement does it follow that $\bigtriangleup ADC \cong \bigtriangleup BCD$ ?

Answer

In addition to the fact that $\overline{AD} \cong \overline{BC}$ and $\overline{AC} \cong \overline{BD}$ , we also have that $\overline{DC} \cong \overline{DC}$ , since, by the Reflexive Property of Congruence, any segment is congruent to itself. We can restate this in a more usable form as $\overline{DC} \cong \overline{CD}$ ; since we have three side congruences between triangles, it follows from the Side-Side-Side Postulate that $\bigtriangleup ADC \cong \bigtriangleup BCD$ .

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ such that

$\overline{AB} \cong \overline{AC}$

$\overline{DE} \cong \overline{DF}$

$\overline{BC} \cong \overline{EF}$

Which statement(s) must be true?

(a) $\bigtriangleup ABC \sim \bigtriangleup DEF$

(b) $\bigtriangleup ABC \cong \bigtriangleup DEF$

Answer

Neither similarity nor congruence of the two triangles follows from the statements given, as can be seen from the figure below:

$\overline{AB} \cong \overline{AC}$ , $\overline{DE} \cong \overline{DF}$ , and $\overline{BC} \cong \overline{EF}$ . However, the triangles are not similar, as

$\frac{EF}{BC} = \frac{10}{10}= 1$

and

$\frac{DE}{AB} = \frac{15}{10} = \frac{3}{2}$

$\frac{DE}{AB} e \frac{EF}{BC}$ , so at least one pair of corresponding sides is not in proportion. Therefore,

$\bigtriangleup ABC sim \bigtriangleup DEF$

The triangles are not similar, and thus cannot be congruent either, so neither statement holds.

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