Congruent Triangles Problems with Solutions

Explore postulates and theorems on congruent triangles using examples and practice problems with detailed solutions.

Side-Angle-Side (SAS) Congruence Postulate

If two sides \(CA\) and \(CB\) and the included angle \(\angle BCA\) of a triangle are congruent to the corresponding two sides \(C'A'\) and \(C'B'\) and the included angle \(\angle B'C'A'\) in another triangle, then the triangles are congruent.

Example 1

Let \(ABCD\) be a parallelogram and \(AC\) be one of its diagonals. What can you say about \(\triangle ABC\) and \(\triangle CDA\)? Explain.

Side-Side-Side (SSS) Congruence Postulate

If the three sides \(AB\), \(BC\), and \(CA\) of a triangle are congruent to the corresponding three sides \(A'B'\), \(B'C'\), and \(C'A'\) of another triangle, then the triangles are congruent.

Example 2

Let \(ABCD\) be a square and \(AC\) be one of its diagonals. What can you say about \(\triangle ABC\) and \(\triangle CDA\)? Explain.

Angle-Side-Angle (ASA) Congruence Postulate

If two angles \(\angle ACB\), \(\angle ABC\) and the included side \(BC\) of a triangle are congruent to the corresponding two angles \(\angle A'C'B'\), \(\angle A'B'C'\) and included side \(B'C'\) in another triangle, then the triangles are congruent.

Example 3

\(\triangle ABC\) is isosceles with \(AB \cong BC\). \(BB'\) is the angle bisector of \(\angle B\). Show that \(\triangle ABB' \cong \triangle CBB'\).

Angle-Angle-Side (AAS) Congruence Theorem

If two angles \(\angle BAC\), \(\angle ACB\) and a non-included side \(AB\) of a triangle are congruent to the corresponding two angles \(\angle B'A'C'\), \(\angle A'C'B'\) and side \(A'B'\) in another triangle, then the triangles are congruent.

Example 4

Hypotenuse-Leg (HL) Congruence Theorem

If the hypotenuse \(BC\) and a leg \(BA\) of a right triangle are congruent to the corresponding hypotenuse \(B'C'\) and leg \(B'A'\) of another right triangle, then the triangles are congruent.

Example 5

Show that the two right triangles below are congruent.

Practice Problems

Problem 1

In isosceles \(\triangle ABC\), \(BA \cong BC\). Points \(M\) and \(N\) are on \(AC\) such that \(MA \cong MB\) and \(NB \cong NC\). Show that \(\triangle AMB \cong \triangle CNB\).

Problem 2

\(ABCD\) is a parallelogram and \(BEFC\) is a square. Show that \(\triangle ABE \cong \triangle DCF\).

Problem 3

\(ABCD\) is a square. \(C'\) is on \(BA\) and \(B'\) is on \(AD\) such that \(BB' \perp CC'\). Show that \(\triangle AB'B \cong \triangle BC'C\).

Problem 4

\(ABC\) is a triangle with midpoint \(M\) of \(AC\). Points \(I\) and \(J\) are on \(BM\) such that \(AI \perp BM\) and \(CJ \perp BM\). Show that \(\triangle AIM \cong \triangle CJM\).

Problem 5

\(\triangle ABC\) is isosceles with \(BA \cong BC\). Points \(K\) on \(AB\) and \(L\) on \(BC\) have perpendiculars \(KK' \perp AC\) and \(LL' \perp AC\) with \(KK' \cong LL'\). Show that \(\triangle KK'M \cong \triangle LL'M\).

Solutions to Examples

Example 1 Solution

In parallelogram \(ABCD\):

1. Opposite sides are congruent: \(BC \cong AD\) and \(AB \cong CD\)
2. Opposite angles are congruent: \(\angle ABC \cong \angle CDA\)
3. Triangles \(ABC\) and \(CDA\) have two congruent sides and the included angle: \(BC \cong AD\), \(AB \cong CD\), and \(\angle ABC \cong \angle CDA\)

Example 2 Solution

In square \(ABCD\):

1. All sides are congruent: \(AB \cong BC \cong CD \cong DA\)
2. Diagonal \(AC\) is common to both triangles
3. Triangles \(ABC\) and \(CDA\) have three congruent sides: \(AB \cong CD\), \(BC \cong DA\), and \(AC \cong CA\)

Example 3 Solution

Since \(\triangle ABC\) is isosceles with \(AB \cong BC\), we have \(\angle A \cong \angle C\).
\(BB'\) bisects \(\angle B\), so \(\angle ABB' \cong \angle CBB'\).
Side \(BB'\) is common to both triangles.
By ASA postulate, \(\triangle ABB' \cong \triangle CBB'\).

Example 4 Solution

In \(\triangle ABC\): \(\angle ABC = 180° - (25° + 125°) = 30°\).
Triangles \(ABC\) and \(PQR\) have:

1. \(\angle ABC \cong \angle QPR = 30°\)
2. \(\angle BAC \cong \angle PQR = 25°\)
3. Side \(BC \cong PR\)

Example 5 Solution

In \(\triangle ABC\), using Pythagorean theorem: \(AB = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\).
Both triangles are right triangles with:

1. Hypotenuse \(BC \cong B'C' = 5\)
2. Leg \(AB \cong A'B' = 4\)

Solutions to Practice Problems

Problem 1 Solution

Since \(MA \cong MB\), \(\triangle AMB\) is isosceles ⇒ \(\angle BAM \cong \angle ABM\).
Since \(NB \cong NC\), \(\triangle CNB\) is isosceles ⇒ \(\angle BCN \cong \angle CBN\).
\(\triangle ABC\) is isosceles ⇒ \(\angle BAM \cong \angle BCN\).
Thus all four angles \(\angle BAM\), \(\angle ABM\), \(\angle CBN\), \(\angle BCN\) are congruent.
Triangles \(AMB\) and \(CNB\) have:

1. \(AB \cong CB\) (isosceles triangle legs)
2. \(\angle BAM \cong \angle BCN\)
3. \(\angle ABM \cong \angle CBN\)

Problem 2 Solution

In parallelogram \(ABCD\): \(BA \cong CD\).
In square \(BEFC\): \(EB \cong FC\).
Angles \(\angle EBA\) and \(\angle FCD\) are congruent (corresponding angles between parallel lines).
Triangles \(ABE\) and \(DCF\) have:

1. \(EB \cong FC\)
2. \(BA \cong CD\)
3. \(\angle EBA \cong \angle FCD\)

Problem 3 Solution

Since \(ABCD\) is a square: \(\angle BAB' \cong \angle CBC' = 90°\).
\(BC \parallel AD\) with transversal \(BB'\) ⇒ \(\angle OBC \cong \angle BB'A\) (alternate interior).
In right \(\triangle CBO\): \(\angle OBC + \angle BCO = 90°\).
In right \(\triangle ABB'\): \(\angle ABB' + \angle BB'A = 90°\).
Since \(\angle OBC \cong \angle BB'A\), then \(\angle ABB' \cong \angle BCC'\).
Triangles \(AB'B\) and \(BC'C\) have:

1. \(BC \cong BA\) (square sides)
2. \(\angle BCC' \cong \angle ABB'\)
3. \(\angle CBC' \cong \angle BAB' = 90°\)

Problem 4 Solution

\(M\) is midpoint of \(AC\) ⇒ \(AM \cong MC\).
\(AI \parallel CJ\) (both perpendicular to \(BM\)).
\(\angle MAI \cong \angle MCJ\) (alternate interior angles).
\(\angle AMI \cong \angle JMC\) (vertical angles).
Triangles \(AIM\) and \(CJM\) have:

1. \(AM \cong MC\)
2. \(\angle MAI \cong \angle MCJ\)
3. \(\angle AMI \cong \angle JMC\)

Problem 5 Solution

\(KK' \parallel LL'\) (both perpendicular to \(AC\)).
\(\angle K'KM \cong \angle L'LM\) (alternate interior angles).
\(\angle KK'M \cong \angle LL'M = 90°\).
Given \(KK' \cong LL'\).
Triangles \(KK'M\) and \(LL'M\) have:

1. \(KK' \cong LL'\)
2. \(\angle K'KM \cong \angle L'LM\)
3. \(\angle KK'M \cong \angle LL'M\)

Additional Resources