Sample Proof on Triangle Congruence Part 3

Recall that for real numbers a, b, and c are real numbers, if a = b and b = c, then a = c. This is called Transitive Property of Equality. This is also the same with congruence. If A, B, and C are polygons, and if A is congruent to B, and B is congruent C, then A is congruent to C. This is called the Transitive Property of Congruence. We will use this to prove the following problem.

Given: \overline{BE} \cong \overline{DC} and \overline{BD} \cong \overline{CA}.

Prove: \triangle DBE \cong \triangle CAB.


It is given that \overline{BE} \cong \overline{DC}.

triangle congruence

Now, by reflexive property, that is a segment is congruent to itself, \overline{BD} \cong \overline{BD}

Notice that \overline{BE} is the hypotenuse of \triangle DBE and \overline{DC} is the hypotenuse of \triangle BDC. Since both are right triangles

\triangle BDC \cong \triangle DBE are congruent by the Hypotenuse Leg Theorem.

Next, we prove that \triangle BDC \cong \triangle CAB.

triangle congruence 2

It is given that \overline{BD} \cong \overline{CA}.

\angle ACB \cong \angle DBC since they are both right angles.

Now, \overline{BC} \cong \overline{BC} by reflexive property.

Therefore, \triangle BDC \cong \triangle CAB by the SAS Congruence Theorem.

Now, since \triangle BDC \cong \triangle DBE, and \triangle BDC \cong CAB,

\triangle DBE \cong \triangle CAB by Transitive Property of Congruence.

This is what we want to prove.

Reference: UCSMP, Geometry

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