given ( riangle bae ) is isosceles with base angles ( angle bea ) and ( angle eba ) and that ( angle aec = 115^circ ), is ( riangle ebc sim riangle aec )? if so, by what criterion?
a yes, by aa criterion
b yes, by sas criterion
c yes, by ssa criterion
d no, not possible to tell.

Question

given ( 	riangle bae ) is isosceles with base angles ( angle bea ) and ( angle eba ) and that ( angle aec = 115^circ ), is ( 	riangle ebc sim 	riangle aec )? if so, by what criterion?
a yes, by aa criterion
b yes, by sas criterion
c yes, by ssa criterion
d no, not possible to tell.

Accepted Answer

# Brief Explanations:
1. First, in isosceles $	riangle BAE$, $\angle EBA = 65^{\circ}$, so $\angle BEA=\angle EBA = 65^{\circ}$ (base angles of isosceles triangle are equal). Then, $\angle BEC = 180^{\circ}-\angle AEC=180 - 115 = 65^{\circ}$ (linear pair).
2. Now, $\angle BEC=\angle BEA = 65^{\circ}$, and $\angle ECB$ and $\angle EAC$: Also, $\angle EBC$ and $\angle AEC$? Wait, let's re - check. $\angle BEC = 65^{\circ}$, $\angle AEC = 115^{\circ}$, and in $	riangle BAE$, $\angle A=180 - 65 - 65 = 50^{\circ}$. Now, for $	riangle EBC$ and $	riangle AEC$: $\angle BEC=\angle EAC$? Wait, no, let's find the angles. $\angle BEC = 65^{\circ}$, $\angle AEC = 115^{\circ}$, and $\angle EBC$: Wait, $\angle EBA = 65^{\circ}$, so $\angle EBC=180 - 65 = 115^{\circ}$? Wait, no, $\angle EBA = 65^{\circ}$, so $\angle EBC = 180^{\circ}-\angle EBA=115^{\circ}$ (linear pair? Wait, no, $\angle EBA$ and $\angle EBC$ are adjacent angles on a straight line? Wait, point B is on AC? Wait, the diagram: A---E---D on a line, B is on AC. So $\angle EBA$ and $\angle EBC$: if B is on AC, then $\angle EBA+\angle EBC = 180^{\circ}$, so $\angle EBC = 180 - 65 = 115^{\circ}$. And $\angle AEC = 115^{\circ}$, so $\angle EBC=\angle AEC = 115^{\circ}$. Also, $\angle BEC=\angle EAC$? Wait, $\angle BEC = 65^{\circ}$, and $\angle EAC$: in $	riangle BAE$, $\angle A=180 - 65 - 65 = 50^{\circ}$? No, wait, $\angle BEC = 65^{\circ}$, and $\angle EAC$: Wait, maybe better to use AA similarity. $\angle BEC=\angle EAC$? Wait, no, let's see: $\angle BEC = 65^{\circ}$, $\angle AEC = 115^{\circ}$, $\angle EBC = 115^{\circ}$, $\angle BEC = 65^{\circ}$, and $\angle ECB$ and $\angle ECA$? Wait, actually, $\angle BEC=\angle EAC$? No, let's do it step by step.

First, calculate $\angle A$ in $	riangle BAE$: $\angle A=180^{\circ}-2	imes65^{\circ}=50^{\circ}$.

$\angle BEC = 180^{\circ}-\angle AEC = 180 - 115 = 65^{\circ}$, so $\angle BEC=\angle A$ (since $\angle A = 50^{\circ}$? No, that's wrong. Wait, I made a mistake. Let's recalculate $\angle BEC$: $\angle AEC = 115^{\circ}$, and $\angle BEA = 65^{\circ}$, so $\angle BEC=\angle AEC-\angle BEA$? No, A---E---D is a straight line, so $\angle AEB+\angle BEC+\angle CED$? No, A---E---D is a straight line, so $\angle AED = 180^{\circ}$, so $\angle AEC+\angle CED = 180^{\circ}$, but E is on AD, so A---E---D, so $\angle AEB+\angle BEC = 180^{\circ}$? Wait, no, $\angle AEC$ is at E, between A and C, and B is on AC. So $\angle BEA$ is at E, between B and A, and $\angle BEC$ is at E, between B and C. So $\angle BEA+\angle BEC=\angle AEC$? No, that can't be. Wait, the problem says $\angle AEC = 115^{\circ}$, and $	riangle BAE$ is isosceles with base angles $\angle BEA$ and $\angle EBA$ (so $BA = BE$? Wait, no, base angles are $\angle BEA$ and $\angle EBA$, so the equal sides are $AB = AE$? Wait, in a triangle, the sides opposite equal angles are equal. So if $\angle BEA=\angle EBA$, then $AB = AE$.

So $\angle BEA = 65^{\circ}$, so $\angle A=180 - 65 - 65 = 50^{\circ}$. Then $\angle BEC=180 - \angle AEC=65^{\circ}$ (since $\angle AEC = 115^{\circ}$, and $\angle AEC+\angle BEC = 180^{\circ}$? Wait, no, A---E---D is a straight line, so $\angle AED = 180^{\circ}$, so $\angle AEC+\angle CED = 180^{\circ}$, but B is on AC, so E is on AD, B is on AC. So $\angle BEA$ is at E, between B and A, $\angle BEC$ is at E, between B and C, and $\angle AEC$ is at E, between A and C. So $\angle BEA+\angle BEC=\angle AEC$? Then $65+\angle BEC = 115$, so $\angle BEC = 50^{\circ}$? Wait, I think I messed up the diagram.

Wait, let's start over. Given $	riangle BAE$ is isosceles with base angles $\angle BEA$ and $\angle EBA$, so $\angle BEA=\angle EBA = 65^{\circ}$, so $\angle A=180 - 65 - 65 = 50^{\circ}$. $\angle AEC = 115^{\circ}$, so $\angle BEC=180 - 115 = 65^{\circ}$ (if E is on AD, and B is on AC, then $\angle AEB+\angle BEC = 180^{\circ}$? No, $\angle AEB$ is $\angle BEA = 65^{\circ}$, so $\angle BEC=180 - 65 = 115^{\circ}$? I'm getting confused. Let's use the AA similarity criterion.

AA similarity: If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar.

In $	riangle EBC$ and $	riangle AEC$:

- $\angle BEC$: Let's find $\angle BEC$. Since $\angle AEC = 115^{\circ}$, and $\angle BEA = 65^{\circ}$, and $\angle BEA+\angle BEC = 180^{\circ}$ (linear pair, because A, E, D are colinear, and B is a point such that E is connected to B and C), so $\angle BEC=180 - 65 = 115^{\circ}$? No, that's not. Wait, $\angle AEC = 115^{\circ}$, so $\angle BEC = 180 - 115 = 65^{\circ}$. And $\angle EBA = 65^{\circ}$, so $\angle EBC = 180 - 65 = 115^{\circ}$ (since $\angle EBA$ and $\angle EBC$ are supplementary). Now, $\angle EBC = 115^{\circ}$ and $\angle AEC = 115^{\circ}$, so $\angle EBC=\angle AEC$. Also, $\angle BEC = 65^{\circ}$ and $\angle EAC$: In $	riangle BAE$, $\angle A=180 - 65 - 65 = 50^{\circ}$? No, that's not. Wait, maybe $\angle BEC=\angle EAC$. Wait, $\angle BEC = 65^{\circ}$, and $\angle EAC$: Let's calculate $\angle EAC$. In $	riangle AEC$, $\angle AEC = 115^{\circ}$, so $\angle EAC+\angle ECA=65^{\circ}$. In $	riangle BAE$, $\angle A = 50^{\circ}$, so $\angle EAC = 50^{\circ}$? No, this is getting too confusing. Let's use the AA criterion.

We know that $\angle BEC=\angle BEA = 65^{\circ}$ (wait, no). Wait, the correct way:

1. In $	riangle BAE$, $\angle EBA = 65^{\circ}$, $\angle BEA = 65^{\circ}$, so $\angle A=50^{\circ}$.

2. $\angle AEC = 115^{\circ}$, so $\angle BEC=180 - 115 = 65^{\circ}$ (linear pair with $\angle AEC$).

3. So $\angle BEC=\angle BEA = 65^{\circ}$.

4. Also, $\angle EBC$: since $\angle EBA = 65^{\circ}$, $\angle EBC = 180 - 65 = 115^{\circ}$ (supplementary angles). And $\angle AEC = 115^{\circ}$, so $\angle EBC=\angle AEC$.

5. Now, in $	riangle EBC$ and $	riangle AEC$: $\angle BEC=\angle EAC$? No, wait, $\angle BEC = 65^{\circ}$, $\angle EAC = 50^{\circ}$? No, I think I made a mistake in angle identification.

Wait, let's look at the angles for AA:

- $\angle BEC$ and $\angle EAC$: No, better: $\angle BEC = 65^{\circ}$, $\angle A = 50^{\circ}$? No. Wait, the correct approach is:

Since $	riangle BAE$ is isosceles with $\angle BEA=\angle EBA = 65^{\circ}$, then $\angle A=180 - 65 - 65 = 50^{\circ}$.

$\angle AEC = 115^{\circ}$, so $\angle BEC=180 - 115 = 65^{\circ}$, so $\angle BEC=\angle BEA = 65^{\circ}$.

Now, $\angle EBC$: $\angle EBA = 65^{\circ}$, so $\angle EBC = 180 - 65 = 115^{\circ}$, and $\angle AEC = 115^{\circ}$, so $\angle EBC=\angle AEC$.

Now, in $	riangle EBC$ and $	riangle AEC$:

- $\angle BEC=\angle EAC$? No, $\angle BEC = 65^{\circ}$, $\angle EAC = 50^{\circ}$? No, I'm wrong. Wait, $\angle BEC = 65^{\circ}$, $\angle A = 50^{\circ}$, so that's not equal. Wait, maybe $\angle BEC=\angle ECA$? No.

Wait, the correct AA: $\angle BEC=\angle BEA = 65^{\circ}$, and $\angle ECB=\angle ECA$? No, maybe I messed up the diagram. Let's think differently.

The key is that $\angle BEC = 65^{\circ}$ (since $\angle AEC = 115^{\circ}$, linear pair) and $\angle BEA = 65^{\circ}$ (base angle of isosceles triangle). Also, $\angle EBC$ and $\angle AEC$: $\angle EBC = 180 - 65 = 115^{\circ}$ (since $\angle EBA = 65^{\circ}$), and $\angle AEC = 115^{\circ}$, so $\angle EBC=\angle AEC$. Then, in $	riangle EBC$ and $	riangle AEC$:

- $\angle BEC=\angle EAC$? No, $\angle EAC = 180 - 65 - 65 = 50^{\circ}$, $\angle BEC = 65^{\circ}$. Wait, no, $\angle BEC = 65^{\circ}$, $\angle AEC = 115^{\circ}$, $\angle EBC = 115^{\circ}$, $\angle BEA = 65^{\circ}$. So $\angle BEC=\angle BEA = 65^{\circ}$, and $\angle EBC=\angle AEC = 115^{\circ}$. So in $	riangle EBC$ and $	riangle AEC$:

$\angle BEC=\angle EAC$? No, $\angle EAC = 50^{\circ}$, $\angle BEC = 65^{\circ}$. Wait, I think I made a mistake in angle assignment. Let's start over with correct angle calculations:

1. In $	riangle BAE$, it is isosceles with base angles $\angle BEA$ and $\angle EBA$, so $\angle BEA=\angle EBA = 65^{\circ}$. Therefore, $\angle A=180^{\circ}-\angle BEA-\angle EBA=180 - 65 - 65 = 50^{\circ}$.

2. $\angle AEC = 115^{\circ}$, so $\angle BEC=180^{\circ}-\angle AEC = 65^{\circ}$ (because $\angle AEC$ and $\angle BEC$ are supplementary as they form a linear pair along line AD? Wait, no, if E is on AD, and B is on AC, then $\angle BEA$ and $\angle BEC$ are adjacent angles at E, with $\angle BEA+\angle BEC=\angle AEC$? No, that would mean $65+\angle BEC = 115$, so $\angle BEC = 50^{\circ}$. Oh! Here's the mistake. I thought $\angle AEC$ and $\angle BEC$ are supplementary, but actually, $\angle BEA+\angle BEC=\angle AEC$. So $\angle BEC=\angle AEC-\angle BEA=115 - 65 = 50^{\circ}$. Then $\angle BEC=\angle A = 50^{\circ}$ (since $\angle A = 50^{\circ}$). Now, $\angle EBC$: in $	riangle BAE$, $\angle EBA = 65^{\circ}$, so $\angle EBC = 180 - 65 = 115^{\circ}$ (supplementary to $\angle EBA$), and $\angle AEC = 115^{\circ}$, so $\angle EBC=\angle AEC$. Now, in $	riangle EBC$ and $	riangle AEC$:

- $\angle BEC=\angle A = 50^{\circ}$

- $\angle EBC=\angle AEC = 115^{\circ}$

So by AA (Angle - Angle) criterion, since two angles of $	riangle EBC$ are equal to two angles of $	riangle AEC$ ($\angle BEC=\angle A$ and $\angle EBC=\angle AEC$), the triangles are similar.

# Answer:
A. yes, by AA criterion

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