Congruent Triangles — Free Geometry Review Games.

SAS (Side-Angle-Side) uses two sides and the angle between them.

SSS (Side-Side-Side) proves congruence when all three sides match.

CPCTC is used after proving triangles congruent to show individual parts are congruent.

An isosceles triangle has at least two sides of equal length.

The three interior angles of any triangle always sum to 180 degrees.

SSA (Side-Side-Angle) does not guarantee congruence (ambiguous case).

Angle C = 180 - 50 - 70 = 60 degrees.

The base angles (opposite the equal sides) of an isosceles triangle are congruent.

AAS (Angle-Angle-Side) uses two angles and a side not between them.

All three angles in an equilateral triangle are 60 degrees (180/3).

2x+3x+4x=180, 9x=180, x=20. Angle B = 3(20) = 60 degrees.

The exterior angle theorem states it equals the sum of the two non-adjacent interior angles.

HL (Hypotenuse-Leg) is a congruence theorem specific to right triangles.

All three sides are equal, so it is equilateral.

Exterior angle = sum of remote interior angles: 130 - 55 = 75 degrees.

ASA stands for Angle-Side-Angle. This postulate states that if two angles and the included side of one triangle are congruent to the corresponding two angles and included side of another triangle, the triangles are congruent. 'Angle-Sum-Angle' is a common distractor but 'Sum' is not part of any congruence postulate.

Congruent triangles are identical in both shape and size — all corresponding sides and all corresponding angles are equal. Similar triangles share the same shape but may differ in size. Two triangles can share the same area or perimeter without being congruent, so those choices alone are insufficient.

The vertex angle of an isosceles triangle is the angle included between the two congruent (equal-length) sides. The two base angles are the congruent angles that sit at either end of the unequal side. The vertex angle can be acute, right, or obtuse — it is not always the largest angle.

ASA (Angle-Side-Angle) requires two pairs of congruent angles and the side that lies between those angles (the included side). SAS uses two sides and the included angle. AAS uses two angles and a non-included side. HL applies only to right triangles using the hypotenuse and one leg.

The Reflexive Property states that any segment, angle, or figure is congruent to itself — for example, \(\overline{AB} \cong \overline{AB}\) and \(\angle A \cong \angle A\). This property is frequently used in triangle congruence proofs when two triangles share a common side or common angle, providing the third congruent pair needed for a postulate.

CPCTC (Corresponding Parts of Congruent Triangles are Congruent) is a conclusion drawn after triangles are proven congruent using a postulate such as SSS, SAS, ASA, AAS, or HL. It cannot appear before that congruence is established. CPCTC applies equally to both corresponding sides and corresponding angles.

The Isosceles Triangle Theorem states that equal sides imply equal opposite angles. Its converse reverses this: if two angles of a triangle are congruent, then the sides opposite those angles are congruent, making the triangle isosceles. The converse does not require the triangle to be equilateral — equilateral triangles are a special case where all three angles are \(60°\).

SAS (Side-Angle-Side) applies when two sides and the angle included between them are congruent in both triangles. Here \(AB = DE\) and \(BC = EF\) are the two sides, and \(\angle B = \angle E\) is the angle between them in each triangle, satisfying SAS. SSS would require a third pair of sides. ASA and AAS both require two pairs of angles, not two pairs of sides.

Since $\triangle PQR \cong \triangle XYZ$, corresponding sides are equal by CPCTC, so \(PQ = XY\). Setting the expressions equal: \(3a - 1 = a + 7 \Rightarrow 2a = 8 \Rightarrow a = 4\). Therefore \(PQ = 3(4) - 1 = 11\). A common error is substituting \(a = 4\) into \(XY = a + 7 = 11\) to verify but forgetting to evaluate \(PQ\) separately.

Since \(AB = AC\), the Isosceles Triangle Theorem guarantees that the base angles $\angle ABC$ and $\angle ACB$ are congruent. Using the Triangle Angle Sum: $40° + 2(\angle ABC) = 180°$, so $2(\angle ABC) = 140°$ and $\angle ABC = 70°$. A common error is assuming $\angle ABC = 40°$ (equal to the vertex angle), which would leave only \(100°\) split between the remaining angle and is incorrect.

ASA requires that the congruent side lies between the two congruent angles — it is the included side. If the congruent side is opposite one of the angles rather than between them, the applicable postulate is AAS, not ASA. Choosing any arbitrary side would not satisfy the strict requirement of ASA. The hypotenuse condition is specific to HL for right triangles.

Once two triangles are proven congruent, CPCTC (Corresponding Parts of Congruent Triangles are Congruent) justifies that any pair of corresponding parts — whether sides or angles — are congruent. The ASA Postulate is a tool used to establish triangle congruence, not to extract conclusions from it. The Reflexive Property applies when a figure is compared to itself.

Both AAS and ASA establish congruence using two pairs of congruent angles and one pair of congruent sides. The critical difference is the position of that side: ASA requires the side to be included (lying between the two congruent angles), while AAS requires a non-included side (not between the two angles). Both are valid for any triangle, not just right triangles.

ASA requires the side included between the two known angles. In $\triangle ABC$, the side between \(\angle A\) and \(\angle B\) is \(\overline{AB}\); in $\triangle PQR$, the side between \(\angle P\) and \(\angle Q\) is \(\overline{PQ}\). So \(AB = PQ\) provides the included side, completing ASA. Choosing \(BC = QR\) would give AAS (a non-included side). Knowing \(\angle C = \angle R\) is automatic from the angle sum and gives only AAA, which proves similarity but not congruence.

AAA establishes that two triangles have the same shape (they are similar), but not necessarily the same size. For instance, a \(30°\)-\(60°\)-\(90°\) triangle with legs \(1\) and \(\sqrt{3}\) and another with legs \(2\) and \(2\sqrt{3}\) have identical angles but different side lengths — they are similar, not congruent. At least one pair of congruent sides is needed to conclude congruence.

The three angles must sum to \(180°\): \((4x) + (2x + 30) + (2x + 30) = 180\). Combining like terms: \(8x + 60 = 180\), so \(8x = 120\) and \(x = 15\). The vertex angle is \(4(15) = 60°\) and each base angle is \(2(15) + 30 = 60°\), confirming this is actually an equilateral triangle — a special case of an isosceles triangle.

We have two pairs of congruent angles (\(\angle A = \angle D\) and \(\angle B = \angle E\)) and one pair of congruent sides (\(BC = EF\)). Side \(\overline{BC}\) is opposite \(\angle A\) in $\triangle ABC$, and \(\overline{EF}\) is opposite \(\angle D\) in $\triangle DEF$ — these sides are not between the two known angles, making this a non-included side. Therefore AAS applies. ASA would require \(\overline{BC}\) to be between \(\angle B\) and \(\angle C\), but \(\angle C\) is not among the given congruent angles.

The HL (Hypotenuse-Leg) Theorem applies exclusively to right triangles: if the hypotenuse and one leg of two right triangles are congruent, the triangles are congruent. Here, both triangles have hypotenuse \(13\) and leg \(5\). The missing leg in each triangle must equal \(\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\), confirming full congruence. SAS, ASA, and AAS would require angle information that is not directly provided.

In $\triangle ABE$ and $\triangle ACD$: \(AB = AC\) (given), $\angle BAE = \angle CAD$ — wait, both triangles share \(\angle A\) as the included angle (Reflexive Property), and \(AE = AD\) (given). The angle \(\angle A\) is included between sides \(AB\) and \(AE\) in $\triangle ABE$, and between sides \(AC\) and \(AD\) in $\triangle ACD$, satisfying SAS. SSS would require knowing \(BE = CD\) first, but that is actually a consequence (provable later by CPCTC), not a given.

By CPCTC, corresponding sides of congruent triangles are equal. For \(AB = DE\): \(3y - 2 = y + 8 \Rightarrow 2y = 10 \Rightarrow y = 5\), so \(AB = 3(5) - 2 = 13\). For \(BC = EF\): \(2z + 1 = z + 5 \Rightarrow z = 4\), so \(BC = 2(4) + 1 = 9\). Therefore \(AB + BC = 13 + 9 = 22\).

Since \(AB = BC\), the angles opposite these equal sides are congruent. Side \(AB\) is opposite $\angle ACB$, and side \(BC\) is opposite $\angle BAC$, so $\angle BAC = \angle ACB = (3x + 5)°$. Using the Triangle Angle Sum: \((3x + 5) + (2x + 10) + (3x + 5) = 180 \Rightarrow 8x + 20 = 180 \Rightarrow x = 20\). Therefore $\angle BAC = 3(20) + 5 = 65°$.

The Transitive Property of Congruence states: if figure \(X \cong\) figure \(Y\) and figure \(Y \cong\) figure \(Z\), then figure \(X \cong\) figure \(Z\). Applying this directly gives $\triangle ABC \cong \triangle GHI$. CPCTC is used to extract specific corresponding-part equalities after congruence is established — it does not establish congruence itself. Since all three triangles are congruent, they are identical in shape and size, not merely similar.

In $\triangle ABM$ and $\triangle ACM$: \(BM = CM\) (M is the midpoint of \(\overline{BC}\)), $\angle AMB = \angle AMC = 90°$ (\(AM \perp BC\)), and \(AM = AM\) (Reflexive Property). The right angle is included between sides \(AM\) and \(BM\) (or \(CM\)), satisfying SAS. HL could seem applicable since both are right triangles, but HL requires congruent hypotenuses (\(AB = AC\)) — this is not given; it is actually a consequence provable afterward via CPCTC.

In $\triangle ABC$ and $\triangle ADC$: $\angle BAC = \angle DAC$ (\(\overline{AC}\) bisects $\angle BAD$), \(AC = AC\) (Reflexive Property), and $\angle BCA = \angle DCA$ (\(\overline{AC}\) bisects $\angle BCD$). Side \(\overline{AC}\) lies between the two pairs of congruent angles, satisfying ASA (Angle-Side-Angle). AAS would require the shared side to be non-included, but here it is clearly the included side.

SSS is justified by \(AB = AC\) (given), \(BD = CD\) (D is the midpoint), and \(AD = AD\) (Reflexive). By CPCTC, corresponding angles $\angle BAD = \angle CAD$, so \(\overline{AD}\) bisects $\angle BAC$. This reveals an important property of isosceles triangles: the median from the apex to the base simultaneously acts as the angle bisector and the perpendicular bisector of the base — all provable through triangle congruence and CPCTC.

The SSS (Side-Side-Side) Postulate guarantees congruence whenever all three corresponding sides of two triangles match in length, since three fixed side lengths determine a unique triangle shape. The choice "SAS Postulate" is wrong because SAS requires two sides and the included angle, not three sides. Students should remember that SSS needs no angle information at all, unlike every other congruence shortcut.

SAS (Side-Angle-Side) requires exactly two pairs of congruent sides with the congruent angle sandwiched between them, since that fixed arrangement forces a unique triangle. The option "Two sides and a non-included angle" is wrong because a non-included angle can produce two different triangles (the ambiguous SSA case), so it does not guarantee congruence. On the exam, always check that the given angle sits physically between the two given sides before labeling a proof step SAS.

HL is a special shortcut that applies exclusively to right triangles, using the hypotenuse and one leg because the right angle guarantees the third side by the Pythagorean relationship. The choice "Any triangle" is incorrect because without a known right angle, matching a hypotenuse-like side and one other side does not fix a unique triangle. Remember that HL is essentially a disguised SSA case that only works because the 90° angle removes the ambiguity.

The Isosceles Triangle Theorem states that angles opposite congruent sides are congruent, so since \(AB = AC\), the angles opposite them, \(\angle C\) and \(\angle B\) respectively, must be equal. The option "\(\angle A\) and \(\angle B\)" is wrong because \(\angle A\) is the vertex angle between the two equal sides, not an angle opposite one of them. Always match each base angle to the side directly across from it when applying this theorem.

The converse reverses the original theorem's hypothesis and conclusion, so proving two angles congruent lets you conclude the sides opposite them are congruent, which is a common proof strategy. The option stating the original (non-converse) theorem about "sides... then the angles" is the forward statement, not its converse. Recognizing which direction of the biconditional you are given determines whether you are proving sides equal from angles or angles equal from sides.

Since \(\angle Y\) is located directly between sides \(XY\) and \(YZ\), and the corresponding parts match in $\triangle LMN$, this fits the SAS Postulate exactly. The distractor "ASA, because two angles and a side are given" is incorrect since only one angle pair, not two, is stated in the given information. When verifying SAS, always trace the diagram to confirm the given angle physically sits between the two given sides.

Since \(DE\) is the side connecting \(\angle D\) and \(\angle E\), and it corresponds to \(RS\) which connects \(\angle R\) and \(\angle S\), the side lies between the two given angles, matching the ASA Postulate. The option "AAS Theorem" is wrong because AAS requires the known side to be non-included (opposite one of the angles), which is not the case here. Before choosing between ASA and AAS, locate exactly where the given side falls relative to the two given angles.

HL requires a right angle, a congruent hypotenuse, and one congruent corresponding leg, so verifying the shared leg genuinely corresponds between the two triangles completes the requirements. The option "The measure of one acute angle" is unnecessary because HL does not require any angle measurements beyond the given right angles. When triangles share a leg, always confirm the shared segment plays the same structural role (leg) in each triangle before invoking HL.

Because $\triangle JKL \cong \triangle MNP$ with \(J\) corresponding to \(M\), CPCTC guarantees that corresponding angles \(\angle J\) and \(\angle M\) are equal once congruence is established. The option citing the "Isosceles Triangle Theorem" is wrong since nothing indicates the triangle is isosceles; the equality here comes purely from the congruence statement. Whenever two triangles are already proven congruent, CPCTC is the correct justification for concluding any pair of corresponding parts are equal.

With two pairs of angles at \(A\) and \(B\) congruent, and the shared side \(\overline{AB}\) lying directly between them by the Reflexive Property, the triangles satisfy ASA. The option "HL Theorem, since a right angle is implied" incorrectly assumes a right angle exists, but no right angle is stated or shown in the given diagram. When two triangles share a side and have congruent angles at both endpoints of that shared side, ASA is almost always the correct tool.

Since all three sides of an equilateral triangle are congruent, the Isosceles Triangle Theorem applies to every pair of sides, forcing all three angles to be congruent, and since they sum to \(180°\), each must equal \(60°\). The option claiming CPCTC alone justifies this is wrong because CPCTC only applies after establishing congruence between two separate triangles, not within a single triangle's own angles. A useful shortcut is that any equilateral triangle is automatically equiangular, with each angle fixed at \(60°\).

SSA is unreliable because, depending on the given lengths and angle, there can be two different triangle configurations (the ambiguous case) that satisfy the same given measurements, so congruence is not guaranteed. The option stating SSA "never produces a valid triangle" is false, since SSA data can and often does produce one or more valid triangles, just not a uniquely determined one. Students should memorize that only SSS, SAS, ASA, AAS, and HL (a special SSA case) reliably prove congruence.

The exterior angle at \(B\) is supplementary to \(\angle B\), so \(\angle B = 180° - 120° = 60°\), and since \(AB = AC\) makes \(\angle B \cong \angle C\) by the Isosceles Triangle Theorem, \(\angle C = 60°\) too, leaving \(\angle A = 180° - 60° - 60° = 60°\). The option "\(120°\)" incorrectly assumes \(\angle A\) equals the exterior angle itself rather than being calculated from the triangle's angle sum. When exterior angles appear with isosceles triangles, always convert to the interior angle first using the supplementary relationship before applying the base angles theorem.

Given \(AB = AC\), the bisected angles $\angle BAD \cong \angle CAD$, and the shared side \(\overline{AD}\) by the Reflexive Property, the two triangles satisfy SAS with \(\angle A\) correctly included between the two given sides. The option citing HL is wrong because no right angle has been established at this stage, so HL cannot yet apply even though it becomes derivable afterward. This configuration is a classic setup showing that the bisector of the vertex angle in an isosceles triangle also becomes the altitude and median, but that follows from CPCTC after the SAS proof, not before.

Since \(PQ = PR\) and \(ST = SU\), and \(PQ = ST\) is given, it follows algebraically that \(PR = SU\) as well, so combined with \(\angle P \cong \angle S\) as the included angle, SAS applies directly. The option claiming the triangles "cannot be proven congruent without knowing \(QR\) and \(TU\)" overlooks that the isosceles conditions supply the missing side equality indirectly. A useful synthesis technique is to use the isosceles property to convert one given side equality into a second one, unlocking SAS where it initially seemed like data was missing.

Since \(\overline{AM}\) is both a median (splitting \(BC\) into equal halves) and an altitude (perpendicular to \(BC\)), the triangles $\triangle ABM$ and $\triangle ACM$ share the right angles at \(M\), the equal segments \(BM = CM\), and the common side \(AM\), giving SAS congruence, so CPCTC yields \(AB = AC\). The option claiming \(BM = AM\) misidentifies corresponding parts, since \(BM\) and \(AM\) are not corresponding sides in the congruence statement $\triangle ABM \cong \triangle ACM$. This problem illustrates the converse idea that a perpendicular median from a vertex forces the triangle to be isosceles at that vertex.

Once $\triangle ABC \cong \triangle ADC$ is established, CPCTC guarantees that all corresponding sides are equal, meaning \(AB\) corresponds to \(AD\) and \(CB\) corresponds to \(CD\), giving both equalities simultaneously. The option stating \(\overline{AC} \perp \overline{BD}\) is unjustified because perpendicularity of the diagonals is not a direct consequence of CPCTC from this particular congruence unless additional information about diagonal \(\overline{BD}\) is given. A key exam skill is listing every corresponding pair from a congruence statement rather than stopping after finding just one useful equality.

Since \(AD = AE\) are given as equal sides of $\triangle ADE$, the Isosceles Triangle Theorem guarantees the angles opposite them, $\angle AED$ and $\angle ADE$, are congruent, making $\triangle ADE$ isosceles in its own right. The option claiming congruence to $\triangle ABC$ by SAS is unsupported because no information ties the lengths \(AD\) or \(AE\) to \(AB\) or \(AC\) specifically enough to establish a full SAS correspondence between the two separate triangles. This setup is a common trick where a smaller isosceles triangle is nested inside a larger one sharing the same vertex angle, and each triangle's isosceles property must be verified independently.

Since \(\overline{BD}\) bisects $\angle ABC$ into two equal angles, is perpendicular to \(\overline{AC}\) giving two right angles at \(D\), and shares side \(\overline{BD}\) with itself, $\triangle ABD \cong \triangle CBD$ by ASA, and CPCTC then yields \(AB = CB\), proving the triangle isosceles. The option claiming the triangle "must be equilateral" is incorrect because equilateral triangles are a special case of isosceles triangles, but nothing in the given information forces all three sides to be equal, only two. This scenario demonstrates the converse principle that when a segment is simultaneously an angle bisector and an altitude from the same vertex, the triangle is guaranteed isosceles at that vertex.

AAA only guarantees that two triangles are similar, not congruent, because matching all three angles fixes the triangle's shape but not its size. For example, 'SAS (Side-Angle-Side)' is a valid postulate because fixing two sides and the included angle between them locks both the shape and size of the triangle uniquely. When checking congruence, always confirm that at least one pair of corresponding sides is included among the given information, since angle-only information can never prove congruence.