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No way. Let $A=$ area. $A = \dfrac{1}{2}ab \sin \theta$ where $\theta$ is the included angle between sides $a$ and $b$. Since $\sin 30^{\circ} = \dfrac{1}{2}= \sin 150^{\circ}$, then a triangle with sides $a,b$ and included angle $30^{\circ}$ has the same area as a triangle with sides $a,b$ and included angle $150^{\circ}$. These two triangles are clearly not congruent.

But your two triangles don't have a common angle.

No way. Let $A=$ area. $A = \dfrac{1}{2}ab \sin \theta$ where $\theta$ is the included angle between sides $a$ and $b$. Since $\sin 30^{\circ} = \dfrac{1}{2}= \sin 150^{\circ}$, then a triangle with sides $a,b$ and included angle $30^{\circ}$ has the same area as a triangle with sides $a,b$ and included angle $150^{\circ}$. These two triangles are clearly not congruent.

There would typically be two distinct possible values for the third side.By the semiperimeter formula, given the area and two sides, the third side can be calculated

I haven't seen the formula of area using the included angle, but it looks like area = (base*height)/2 for an included angle of 90 degrees is a specific case of that just like the Pythagorean Theorem is a special case of the law of cosines. I took the square of the area formula, changed theta to c, and expressed (sin c)^2 as 1-(cos c)^2 to use the same terms as the law of cosines. Then can the law of cosines and area formula with two sides and the included angle be used to derive the semiperimeter formula?

No way. Let $A=$ area. $A = \dfrac{1}{2}ab \sin \theta$ where $\theta$ is the included angle between sides $a$ and $b$. Since $\sin 30^{\circ} = \dfrac{1}{2}= \sin 150^{\circ}$, then a triangle with sides $a,b$ and included angle $30^{\circ}$ has the same area as a triangle with sides $a,b$ and included angle $150^{\circ}$. These two triangles are clearly not congruent.

Take a 3-4-5 right triangle with area of 6. Let's say you have the area and two sides and try to solve for the third side. First, let's say you know the sides of 3 and 4 but don't know they're perpendicular to be the base and height. The square of the area is 36. Let the semiperimeter be p, so p = (a+b+c)/2, a is 3, and b is 4. p = (7+c)/2. p-a = (7+c)/2 - 3. p-b = (7+c)/2 - 4. p-c = (7+c)/2 - c. You know the product of those four terms. The only way I can think of for the triangles to not be congruent is for it to produce an equation of a higher degree that makes multiple solutions. Is that what happens? If a triangle has area of 6 and sides of 3 and 4, what length other than 5 could the other side have?

I don't understand. One side and three angles proves congruence ASA or AAS.Two sides and three angles don't even guarantee congruent triangles.

That made me think of the drawing that shows two possible triangles if you know ASS, which doesn't prove congruence. However, my example is different because I'm talking about equal areas, and the example showing that ASS doesn't prove congruence uses triangles of unequal areas.There would typically be two distinct possible values for the third side.

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It is a standard high school trig area formula. Google it.I haven't seen the formula of area using the included angle,

In the attached picture, $\triangle ABC$ and $\triangle ABD$ both have $AB=AB$ and $BC=BD$ and both have area $\dfrac{1}{2}(8)(3)=12$.

$\triangle DEF$ has sides $8,12,18$ and $\triangle PQR$ has sides $12,18,27$. We have $12=12$, $18=18$ and since the triangles are similar all their angles are the same.I don't understand. One side and three angles proves congruence ASA or AAS.

It is difficult to see who is agreeing with whom in this thread. So I will throw in my additional 2 cents worth. Consider the following picture:

Let's say the original triangle is ABC, and the base AB is equal to the base of the second triangle. Place the vertex C above the base, which sets the height h. Place the second triangle with its equal base on AB and its third vertex above. In order to have the same area its vertex must be on the dotted horizontal line to have the same height. I have indicated the given angle at A in a green arc. The OP's question said nothing about the angles being corresponding angles. It's pretty clear if the second triangle's angle A is the same, the triangles are the same. Consider the case when the second triangle's equal angle is at B, so it is ABC' (the red triangle). Then ABC is congruent to BAC' by symmetry. But what if the equal angle in the second triangle is at the vertex on the dotted line? Look at triangle ABC'' in blue. Clearly the angle at C'' is less than the angle at A. Now look at triangle ABC''' in violet. In the figure the angle at C''' is slightly larger than a right angle and clearly larger than the angle at A. At C'' the angle is less, at C''' the angle is more, so there must be a point X between them (the brown triangle) where the angle at X equals the angle at A. So triangle ABX has an equal angle, an equal side, and the same area. From the sketch they clearly aren't congruent because ABX has an obtuse angle at B and ABC has no obtuse angles.

Let's say the original triangle is ABC, and the base AB is equal to the base of the second triangle. Place the vertex C above the base, which sets the height h. Place the second triangle with its equal base on AB and its third vertex above. In order to have the same area its vertex must be on the dotted horizontal line to have the same height. I have indicated the given angle at A in a green arc. The OP's question said nothing about the angles being corresponding angles. It's pretty clear if the second triangle's angle A is the same, the triangles are the same. Consider the case when the second triangle's equal angle is at B, so it is ABC' (the red triangle). Then ABC is congruent to BAC' by symmetry. But what if the equal angle in the second triangle is at the vertex on the dotted line? Look at triangle ABC'' in blue. Clearly the angle at C'' is less than the angle at A. Now look at triangle ABC''' in violet. In the figure the angle at C''' is slightly larger than a right angle and clearly larger than the angle at A. At C'' the angle is less, at C''' the angle is more, so there must be a point X between them (the brown triangle) where the angle at X equals the angle at A. So triangle ABX has an equal angle, an equal side, and the same area. From the sketch they clearly aren't congruent because ABX has an obtuse angle at B and ABC has no obtuse angles.

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