How to find if of acute / obtuse isosceles triangle are similar - Geometry

Card 0 of 20

Question

Refer to the above diagram. .

True or false: From the information given, it follows that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Answer

By the Angle-Angle Similarity Postulate, if two pairs of corresponding angles of a triangle are congruent, the triangles themselves are similar.

$\angle AVB$ and $\angle CVD$ are a pair of vertical angles, having the same vertex and having sides opposite each other. As such, $\angle AVB \cong \angle CVD$ .

$\angle ABV$ and $\angle CDV$ are alternating interior angles formed by two parallel lines and cut by a transversal . As a consequence, $\angle ABV \cong \angle CDV$ .

The conditions of the Angle-Angle Similarity Postulate are satisfied, and it holds that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Compare your answer with the correct one above

Question

$\bigtriangleup ABC$ and $\bigtriangleup DEF$ are both isosceles triangles;

$m \angle B = 120 ^{\circ }$

$m \angle F = 30 ^{\circ }$

True or false: from the given information, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

Answer

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond respectively to , , and .

$\bigtriangleup DEF$ is an isosceles triangle, so it must have two congruent angles. $\angle F$ has measure $30^{\circ }$ , so either $\angle D$ has this measure, $\angle E$ has this measure, or $\angle D \cong \angle E$ . If we examine the second case, it immediately follows that $\angle B \ncong \angle E$ . One condition of the similarity of triangles is that all pairs of corresponding angles be congruent; since there is at least one case that violates this condition, it does not necessarily follow that $\bigtriangleup ABC \sim \bigtriangleup DEF$ . This makes the correct response "false".

Compare your answer with the correct one above

Question

$\bigtriangleup ABC$ is an equilateral triangle; $\bigtriangleup DEF$ is an equiangular triangle.

True or false: From the given information, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

Answer

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond respectively to , , and .

A triangle is equilateral (having three sides of the same length) if and only if it is also equiangular (having three angles of the same measure, each of which is $60^{\circ }$ ). It follows that all angles of both triangles measure $60^{\circ }$ .

Specifically, $\angle A \cong \angle D$ and $\angle B \cong \angle E$ , making two pairs of corresponding angles congruent. By the Angle-Angle Similarity Postulate, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ , making the correct answer "true".

Compare your answer with the correct one above

Question

Refer to the above diagram. $\angle ABV \cong \angle DCV$ .

True or false: From the information given, it follows that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Answer

The given information is actually inconclusive.

By the Angle-Angle Similarity Postulate, if two pairs of corresponding angles of a triangle are congruent, the triangles themselves are similar. Therefore, we seek to prove two of the following three angle congruence statements:

$\angle AVB \cong \angle CVD$

$\angle ABV \cong \angle CDV$

$\angle BAV \cong \angle DCV$

$\angle AVB$ and $\angle CVD$ are a pair of vertical angles, having the same vertex and having sides opposite each other. As such, $\angle AVB \cong \angle CVD$ .

$\angle ABV \cong \angle DCV$ , but this is not one of the statements we need to prove. Also, without further information - for example, whether and are parallel, which is not given to us - we have no way to prove either of the other two necessary statements.

The correct response is "false".

Compare your answer with the correct one above

Question

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

$\angle A \cong \angle B$

$\angle C \cong \angle F$

$\angle D \cong \angle E$

Which statement(s) must be true?

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

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

Answer

The sum of the measures of the interior angles of a triangle is $180^{\circ }$ , so

$m \angle A + m \angle B + m \angle C = 180^{\circ }$

$m \angle A + m \angle B + m \angle C - m \angle C = 180^{\circ } - m \angle C$

$m \angle A + m \angle B = 180^{\circ } - m \angle C$

Also,

$m \angle A = m \angle B$ ,

so

$m \angle A + m \angle A = 180^{\circ } - m \angle C$

$2 \cdot m \angle A = 180^{\circ } - m \angle C$

$\frac{2 \cdot m \angle A }{2}= \frac{180^{\circ } - m \angle C}{2}$

$m \angle A = \frac{180^{\circ } - m \angle C}{2}$

By similar reasoning, it holds that

$m \angle D = \frac{180^{\circ } - m \angle F}{2}$

Since $m \angle F = m \angle C$ , by substitution,

$m \angle D = \frac{180^{\circ } - m \angle C}{2}$

Therefore,

$m \angle A = m \angle D$ ,

or $\angle A \cong \angle D$

This, along with the statement that $\angle C \cong \angle F$ , sets up the conditions of the Angle-Angle Similarity Postulate - if two angles of one triangle are congruent to the two corresponding angles of another triangle, the two triangles are similar. It follows that

$\bigtriangleup ABC \sim \bigtriangleup DEF$ .

However, congruence cannot be proved, since at least one side congruence is needed to prove this. This is not given in the problem.

Therefore, statement (a) must hold, but not necessarily statement (b).

Compare your answer with the correct one above

Question

$\bigtriangleup ABC$ and $\bigtriangleup DEF$ are both isosceles triangles;

$m \angle B = 120 ^{\circ }$

$m \angle F = 30 ^{\circ }$

True or false: from the given information, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

Answer

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond respectively to , , and .

$\bigtriangleup DEF$ is an isosceles triangle, so it must have two congruent angles. $\angle F$ has measure $30^{\circ }$ , so either $\angle D$ has this measure, $\angle E$ has this measure, or $\angle D \cong \angle E$ . If we examine the second case, it immediately follows that $\angle B \ncong \angle E$ . One condition of the similarity of triangles is that all pairs of corresponding angles be congruent; since there is at least one case that violates this condition, it does not necessarily follow that $\bigtriangleup ABC \sim \bigtriangleup DEF$ . This makes the correct response "false".

Compare your answer with the correct one above

Question

$\bigtriangleup ABC$ is an equilateral triangle; $\bigtriangleup DEF$ is an equiangular triangle.

True or false: From the given information, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

Answer

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond respectively to , , and .

A triangle is equilateral (having three sides of the same length) if and only if it is also equiangular (having three angles of the same measure, each of which is $60^{\circ }$ ). It follows that all angles of both triangles measure $60^{\circ }$ .

Specifically, $\angle A \cong \angle D$ and $\angle B \cong \angle E$ , making two pairs of corresponding angles congruent. By the Angle-Angle Similarity Postulate, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ , making the correct answer "true".

Compare your answer with the correct one above

Question

Refer to the above diagram. .

True or false: From the information given, it follows that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Answer

By the Angle-Angle Similarity Postulate, if two pairs of corresponding angles of a triangle are congruent, the triangles themselves are similar.

$\angle AVB$ and $\angle CVD$ are a pair of vertical angles, having the same vertex and having sides opposite each other. As such, $\angle AVB \cong \angle CVD$ .

$\angle ABV$ and $\angle CDV$ are alternating interior angles formed by two parallel lines and cut by a transversal . As a consequence, $\angle ABV \cong \angle CDV$ .

The conditions of the Angle-Angle Similarity Postulate are satisfied, and it holds that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Compare your answer with the correct one above

Question

Refer to the above diagram. $\angle ABV \cong \angle DCV$ .

True or false: From the information given, it follows that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Answer

The given information is actually inconclusive.

By the Angle-Angle Similarity Postulate, if two pairs of corresponding angles of a triangle are congruent, the triangles themselves are similar. Therefore, we seek to prove two of the following three angle congruence statements:

$\angle AVB \cong \angle CVD$

$\angle ABV \cong \angle CDV$

$\angle BAV \cong \angle DCV$

$\angle AVB$ and $\angle CVD$ are a pair of vertical angles, having the same vertex and having sides opposite each other. As such, $\angle AVB \cong \angle CVD$ .

$\angle ABV \cong \angle DCV$ , but this is not one of the statements we need to prove. Also, without further information - for example, whether and are parallel, which is not given to us - we have no way to prove either of the other two necessary statements.

The correct response is "false".

Compare your answer with the correct one above

Question

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

$\angle A \cong \angle B$

$\angle C \cong \angle F$

$\angle D \cong \angle E$

Which statement(s) must be true?

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

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

Answer

The sum of the measures of the interior angles of a triangle is $180^{\circ }$ , so

$m \angle A + m \angle B + m \angle C = 180^{\circ }$

$m \angle A + m \angle B + m \angle C - m \angle C = 180^{\circ } - m \angle C$

$m \angle A + m \angle B = 180^{\circ } - m \angle C$

Also,

$m \angle A = m \angle B$ ,

so

$m \angle A + m \angle A = 180^{\circ } - m \angle C$

$2 \cdot m \angle A = 180^{\circ } - m \angle C$

$\frac{2 \cdot m \angle A }{2}= \frac{180^{\circ } - m \angle C}{2}$

$m \angle A = \frac{180^{\circ } - m \angle C}{2}$

By similar reasoning, it holds that

$m \angle D = \frac{180^{\circ } - m \angle F}{2}$

Since $m \angle F = m \angle C$ , by substitution,

$m \angle D = \frac{180^{\circ } - m \angle C}{2}$

Therefore,

$m \angle A = m \angle D$ ,

or $\angle A \cong \angle D$

This, along with the statement that $\angle C \cong \angle F$ , sets up the conditions of the Angle-Angle Similarity Postulate - if two angles of one triangle are congruent to the two corresponding angles of another triangle, the two triangles are similar. It follows that

$\bigtriangleup ABC \sim \bigtriangleup DEF$ .

However, congruence cannot be proved, since at least one side congruence is needed to prove this. This is not given in the problem.

Therefore, statement (a) must hold, but not necessarily statement (b).

Compare your answer with the correct one above

Question

$\bigtriangleup ABC$ and $\bigtriangleup DEF$ are both isosceles triangles;

$m \angle B = 120 ^{\circ }$

$m \angle F = 30 ^{\circ }$

True or false: from the given information, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

Answer

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond respectively to , , and .

$\bigtriangleup DEF$ is an isosceles triangle, so it must have two congruent angles. $\angle F$ has measure $30^{\circ }$ , so either $\angle D$ has this measure, $\angle E$ has this measure, or $\angle D \cong \angle E$ . If we examine the second case, it immediately follows that $\angle B \ncong \angle E$ . One condition of the similarity of triangles is that all pairs of corresponding angles be congruent; since there is at least one case that violates this condition, it does not necessarily follow that $\bigtriangleup ABC \sim \bigtriangleup DEF$ . This makes the correct response "false".

Compare your answer with the correct one above

Question

$\bigtriangleup ABC$ is an equilateral triangle; $\bigtriangleup DEF$ is an equiangular triangle.

True or false: From the given information, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

Answer

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond respectively to , , and .

A triangle is equilateral (having three sides of the same length) if and only if it is also equiangular (having three angles of the same measure, each of which is $60^{\circ }$ ). It follows that all angles of both triangles measure $60^{\circ }$ .

Specifically, $\angle A \cong \angle D$ and $\angle B \cong \angle E$ , making two pairs of corresponding angles congruent. By the Angle-Angle Similarity Postulate, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ , making the correct answer "true".

Compare your answer with the correct one above

Question

Refer to the above diagram. .

True or false: From the information given, it follows that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Answer

By the Angle-Angle Similarity Postulate, if two pairs of corresponding angles of a triangle are congruent, the triangles themselves are similar.

$\angle AVB$ and $\angle CVD$ are a pair of vertical angles, having the same vertex and having sides opposite each other. As such, $\angle AVB \cong \angle CVD$ .

$\angle ABV$ and $\angle CDV$ are alternating interior angles formed by two parallel lines and cut by a transversal . As a consequence, $\angle ABV \cong \angle CDV$ .

The conditions of the Angle-Angle Similarity Postulate are satisfied, and it holds that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Compare your answer with the correct one above

Question

Refer to the above diagram. $\angle ABV \cong \angle DCV$ .

True or false: From the information given, it follows that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Answer

The given information is actually inconclusive.

By the Angle-Angle Similarity Postulate, if two pairs of corresponding angles of a triangle are congruent, the triangles themselves are similar. Therefore, we seek to prove two of the following three angle congruence statements:

$\angle AVB \cong \angle CVD$

$\angle ABV \cong \angle CDV$

$\angle BAV \cong \angle DCV$

$\angle AVB$ and $\angle CVD$ are a pair of vertical angles, having the same vertex and having sides opposite each other. As such, $\angle AVB \cong \angle CVD$ .

$\angle ABV \cong \angle DCV$ , but this is not one of the statements we need to prove. Also, without further information - for example, whether and are parallel, which is not given to us - we have no way to prove either of the other two necessary statements.

The correct response is "false".

Compare your answer with the correct one above

Question

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

$\angle A \cong \angle B$

$\angle C \cong \angle F$

$\angle D \cong \angle E$

Which statement(s) must be true?

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

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

Answer

The sum of the measures of the interior angles of a triangle is $180^{\circ }$ , so

$m \angle A + m \angle B + m \angle C = 180^{\circ }$

$m \angle A + m \angle B + m \angle C - m \angle C = 180^{\circ } - m \angle C$

$m \angle A + m \angle B = 180^{\circ } - m \angle C$

Also,

$m \angle A = m \angle B$ ,

so

$m \angle A + m \angle A = 180^{\circ } - m \angle C$

$2 \cdot m \angle A = 180^{\circ } - m \angle C$

$\frac{2 \cdot m \angle A }{2}= \frac{180^{\circ } - m \angle C}{2}$

$m \angle A = \frac{180^{\circ } - m \angle C}{2}$

By similar reasoning, it holds that

$m \angle D = \frac{180^{\circ } - m \angle F}{2}$

Since $m \angle F = m \angle C$ , by substitution,

$m \angle D = \frac{180^{\circ } - m \angle C}{2}$

Therefore,

$m \angle A = m \angle D$ ,

or $\angle A \cong \angle D$

This, along with the statement that $\angle C \cong \angle F$ , sets up the conditions of the Angle-Angle Similarity Postulate - if two angles of one triangle are congruent to the two corresponding angles of another triangle, the two triangles are similar. It follows that

$\bigtriangleup ABC \sim \bigtriangleup DEF$ .

However, congruence cannot be proved, since at least one side congruence is needed to prove this. This is not given in the problem.

Therefore, statement (a) must hold, but not necessarily statement (b).

Compare your answer with the correct one above

Question

Refer to the above diagram. .

True or false: From the information given, it follows that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Answer

By the Angle-Angle Similarity Postulate, if two pairs of corresponding angles of a triangle are congruent, the triangles themselves are similar.

$\angle AVB$ and $\angle CVD$ are a pair of vertical angles, having the same vertex and having sides opposite each other. As such, $\angle AVB \cong \angle CVD$ .

$\angle ABV$ and $\angle CDV$ are alternating interior angles formed by two parallel lines and cut by a transversal . As a consequence, $\angle ABV \cong \angle CDV$ .

The conditions of the Angle-Angle Similarity Postulate are satisfied, and it holds that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Compare your answer with the correct one above

Question

$\bigtriangleup ABC$ and $\bigtriangleup DEF$ are both isosceles triangles;

$m \angle B = 120 ^{\circ }$

$m \angle F = 30 ^{\circ }$

True or false: from the given information, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

Answer

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond respectively to , , and .

$\bigtriangleup DEF$ is an isosceles triangle, so it must have two congruent angles. $\angle F$ has measure $30^{\circ }$ , so either $\angle D$ has this measure, $\angle E$ has this measure, or $\angle D \cong \angle E$ . If we examine the second case, it immediately follows that $\angle B \ncong \angle E$ . One condition of the similarity of triangles is that all pairs of corresponding angles be congruent; since there is at least one case that violates this condition, it does not necessarily follow that $\bigtriangleup ABC \sim \bigtriangleup DEF$ . This makes the correct response "false".

Compare your answer with the correct one above

Question

$\bigtriangleup ABC$ is an equilateral triangle; $\bigtriangleup DEF$ is an equiangular triangle.

True or false: From the given information, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

Answer

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond respectively to , , and .

A triangle is equilateral (having three sides of the same length) if and only if it is also equiangular (having three angles of the same measure, each of which is $60^{\circ }$ ). It follows that all angles of both triangles measure $60^{\circ }$ .

Specifically, $\angle A \cong \angle D$ and $\angle B \cong \angle E$ , making two pairs of corresponding angles congruent. By the Angle-Angle Similarity Postulate, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ , making the correct answer "true".

Compare your answer with the correct one above

Question

Refer to the above diagram. $\angle ABV \cong \angle DCV$ .

True or false: From the information given, it follows that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Answer

The given information is actually inconclusive.

By the Angle-Angle Similarity Postulate, if two pairs of corresponding angles of a triangle are congruent, the triangles themselves are similar. Therefore, we seek to prove two of the following three angle congruence statements:

$\angle AVB \cong \angle CVD$

$\angle ABV \cong \angle CDV$

$\angle BAV \cong \angle DCV$

$\angle AVB$ and $\angle CVD$ are a pair of vertical angles, having the same vertex and having sides opposite each other. As such, $\angle AVB \cong \angle CVD$ .

$\angle ABV \cong \angle DCV$ , but this is not one of the statements we need to prove. Also, without further information - for example, whether and are parallel, which is not given to us - we have no way to prove either of the other two necessary statements.

The correct response is "false".

Compare your answer with the correct one above

Question

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

$\angle A \cong \angle B$

$\angle C \cong \angle F$

$\angle D \cong \angle E$

Which statement(s) must be true?

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

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

Answer

The sum of the measures of the interior angles of a triangle is $180^{\circ }$ , so

$m \angle A + m \angle B + m \angle C = 180^{\circ }$

$m \angle A + m \angle B + m \angle C - m \angle C = 180^{\circ } - m \angle C$

$m \angle A + m \angle B = 180^{\circ } - m \angle C$

Also,

$m \angle A = m \angle B$ ,

so

$m \angle A + m \angle A = 180^{\circ } - m \angle C$

$2 \cdot m \angle A = 180^{\circ } - m \angle C$

$\frac{2 \cdot m \angle A }{2}= \frac{180^{\circ } - m \angle C}{2}$

$m \angle A = \frac{180^{\circ } - m \angle C}{2}$

By similar reasoning, it holds that

$m \angle D = \frac{180^{\circ } - m \angle F}{2}$

Since $m \angle F = m \angle C$ , by substitution,

$m \angle D = \frac{180^{\circ } - m \angle C}{2}$

Therefore,

$m \angle A = m \angle D$ ,

or $\angle A \cong \angle D$

This, along with the statement that $\angle C \cong \angle F$ , sets up the conditions of the Angle-Angle Similarity Postulate - if two angles of one triangle are congruent to the two corresponding angles of another triangle, the two triangles are similar. It follows that

$\bigtriangleup ABC \sim \bigtriangleup DEF$ .

However, congruence cannot be proved, since at least one side congruence is needed to prove this. This is not given in the problem.

Therefore, statement (a) must hold, but not necessarily statement (b).

Compare your answer with the correct one above

Tap the card to reveal the answer