The Encyclopedia of Geometry (0036)

Problem

There are $△ABC$ and $△A’B’C’$.

$$AB=A’B’, \ BC=B’C’, \ and \ ∠A=∠A’.$$

Then, under which of the following conditions can we always have $△ABC≡△A’B’C’$?

(1) Either $△ABC$ or $△A’B’C’$ is an obtuse triangle.

(2) $∠C$ is a right angle.

(3) $BC>AB$.

(4) $AC$ and $A’C’$ are the maximum sides of $△ABC$ and $△A’B’C’$, respectively.

(5) $AB>AC$ and $A’B’>A’C’$.

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Solution

Since the two sets of sides of the two triangles and the angles to one set of them are equal, from the previous question $0035$, the angles to the other set of equal sides are either equal or supplementary to each other. Therefore,

(1) If one supplementary angle is obtuse, the other is acute. Thus, the two triangles are not necessarily congruent.

(2) If one supplementary angle is a right angle, the other angle is also a right angle. Therefore, two triangles are congruent.

(3) If $BC$ is longer than $AB$, then $∠A$ is greater than $∠C$. Also, by definition, $B’C’$ is longer than $A’B’$. Thus, $∠A’$ is greater than $∠C’$. Therefore, both $∠C$ and $∠C’$ are always acute angles, and they are never supplementary to each other. Therefore, the two triangles are congruent.

(4) Since $AC$ and $A’C’$ are the largest sides of each triangle, $∠B$ and $∠B’$ are the largest angles. Therefore, both $∠C$ and $∠C’$ are always less than a right angle, and they are never supplementary to each other. Thus, the two triangles are congruent.

(5) From the previous question $0035$, it is not necessarily congruent.

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Reference

Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2^nd edition), Seikyo-Shinsha, p.11