Given that triangle ABC is congruent to triangle PQR, BAC = QPR, then side BC is equal to side ______ in triangle PQR.

The question states that triangle ABC is congruent to triangle PQR ($\triangle ABC \cong \triangle PQR$).
When two triangles are congruent, their corresponding sides and corresponding angles are equal.
The order of the vertices in the congruence statement indicates which parts correspond.
A corresponds to P.
B corresponds to Q.
C corresponds to R.

We are given that $\angle BAC = \angle QPR$. This is consistent with the congruence statement, as $\angle BAC$ is the angle at vertex A, and $\angle QPR$ is the angle at vertex P.
The question asks which side BC is equal to in triangle PQR.
Since B corresponds to Q and C corresponds to R, the side BC in $\triangle ABC$ corresponds to side QR in $\triangle PQR$.
Therefore, side BC is equal to side QR.
The answer is $\boxed{QR}$.

The image shows two triangles, $\triangle PQR$ and $\triangle TUV$. We need to show that they are congruent.
Let's examine the markings on the triangles:
In $\triangle PQR$:

• Side PQ has two tick marks.
• Side PR has one tick mark.
• Angle Q is marked with a right angle symbol ($90^\circ$).

In $\triangle TUV$:

• Side TU has two tick marks.
• Side TV has one tick mark.
• Angle U is marked with a right angle symbol ($90^\circ$).

Comparing the corresponding parts:

1. Side PQ has two tick marks, and side TU has two tick marks. So, $PQ = TU$.
2. Angle Q is a right angle, and angle U is a right angle. So, $\angle Q = \angle U = 90^\circ$.
3. Side PR has one tick mark, and side TV has one tick mark. So, $PR = TV$.

We have two sides and a non-included angle (SSA). However, for right-angled triangles, if the hypotenuse and one leg are equal, the triangles are congruent by the Hypotenuse-Leg (HL) congruence theorem.
In $\triangle PQR$, PR is the hypotenuse (opposite the right angle Q). PQ is a leg.
In $\triangle TUV$, TV is the hypotenuse (opposite the right angle U). TU is a leg.

We have:

• Hypotenuse $PR = TV$ (both have one tick mark).
• Leg $PQ = TU$ (both have two tick marks).
• Both triangles are right-angled at Q and U respectively.

Therefore, $\triangle PQR \cong \triangle TUV$ by the Hypotenuse-Leg (HL) congruence theorem.
The answer is $\boxed{\text{HL Congruence Theorem}}$.

The question states that $\triangle ABE$ is congruent to $\triangle BDC$ ($\triangle ABE \cong \triangle BDC$).
We are given the following lengths:
AE = 5 cm
EB = 3 cm
AB = 7 cm

We need to find the measure of side BC.
Since the triangles are congruent, their corresponding sides are equal.
From the congruence statement $\triangle ABE \cong \triangle BDC$:

• Side AB corresponds to side BD. So, $AB = BD$.
• Side BE corresponds to side DC. So, $BE = DC$.
• Side AE corresponds to side BC. So, $AE = BC$.

We are given $AE = 5$ cm.
Since $AE = BC$, then $BC = 5$ cm.
The measure of side BC is $\boxed{5 \text{ cm}}$.