Criteria for Congruence of Triangles Questions in English

(N/A) To prove that diagonal $PR$ bisects $\angle QPS$ and $\angle QRS$,we consider the two triangles $\triangle PQR$ and $\triangle PSR$.
$1$. In $\triangle PQR$ and $\triangle PSR$:
- $PQ = PS$ (Given)
- $QR = RS$ (Given)
- $PR = PR$ (Common side)
$2$. By the $SSS$ (Side-Side-Side) congruence criterion,$\triangle PQR \cong \triangle PSR$.
$3$. Since the triangles are congruent,their corresponding parts are equal $(CPCT)$:
- $\angle QPR = \angle SPR$. This implies that $PR$ bisects $\angle QPS$.
- $\angle QRP = \angle SRP$. This implies that $PR$ bisects $\angle QRS$.
Thus,it is proved that diagonal $PR$ bisects both $\angle QPS$ and $\angle QRS$.

(N/A) $1$. Consider the two triangles $\triangle PQR$ and $\triangle PSR$.
$2$. In these triangles,we are given that $PQ = PS$ (Side).
$3$. We are also given that $RQ = RS$ (Side).
$4$. The side $PR$ is common to both triangles,so $PR = PR$ (Side).
$5$. By the $SSS$ (Side-Side-Side) congruence criterion,$\triangle PQR \cong \triangle PSR$.
$6$. Since the triangles are congruent,their corresponding parts are equal ($CPCT$ - Corresponding Parts of Congruent Triangles).
$7$. Therefore,$\angle PQR = \angle PSR$.

(N/A) $1$. Consider the right-angled triangles $\Delta PLQ$ and $\Delta QMP$.
$2$. In these triangles,$PL = QM$ (given),$PQ = QP$ (common side),and $\angle PLQ = \angle QMP = 90^{\circ}$ (altitudes).
$3$. By $RHS$ congruence criterion,$\Delta PLQ \cong \Delta QMP$.
$4$. By $CPCT$,$\angle PQL = \angle QPM$,which implies $\angle PQR = \angle QPR$.
$5$. Since the base angles are equal,the sides opposite to them are equal,so $PR = QR$.
$6$. Similarly,by considering $\Delta QMR$ and $\Delta RNP$,we can prove $\Delta QMR \cong \Delta RNP$,which implies $\angle QRM = \angle RPN$,leading to $PQ = QR$.
$7$. Since $PR = QR$ and $PQ = QR$,we have $PQ = QR = PR$.
$8$. Therefore,$\Delta PQR$ is an equilateral triangle.

(N/A) $1$. In $\Delta PQR$,let $PL \perp QR$,$QM \perp PR$,and $RN \perp PQ$. Given $PL = QM = RN$.
$2$. Consider $\Delta QRN$ and $\Delta RQM$. We have $\angle QNR = \angle RMQ = 90^{\circ}$,$QR = RQ$ (common side),and $RN = QM$ (given).
$3$. By $RHS$ congruence,$\Delta QRN \cong \Delta RQM$. Thus,$\angle RQN = \angle QRM$,which implies $\angle RQP = \angle QRP$.
$4$. Similarly,by considering $\Delta PQL$ and $\Delta QPM$,we can show $\angle QPR = \angle PQR$.
$5$. Since $\angle PQR = \angle QRP$ and $\angle QPR = \angle PQR$,it follows that $\angle PQR = \angle QRP = \angle QPR$.
$6$. Since all angles are equal,$\Delta PQR$ is an equilateral triangle.

(A) $(1)$ False. According to the theorem,in any triangle,the angle opposite to the longer side is larger. Since $XY > XZ$,the angle opposite to $XY$ (which is $\angle Z$) must be greater than the angle opposite to $XZ$ (which is $\angle Y$). Therefore,$\angle Z > \angle Y$.
$(2)$ True. Given $\frac{AB}{PR} = \frac{BC}{QP} = \frac{CA}{RQ} = 1$,it implies $AB = PR$,$BC = QP$,and $CA = RQ$. By the $SSS$ (Side-Side-Side) congruence criterion,$\Delta ABC \cong \Delta RPQ$.

(B) Given that in $\Delta ABC$ and $\Delta PQR$:
$1$. $AB = PR$
$2$. $BC = RQ$
$3$. $\angle B = \angle R$
By the $SAS$ (Side-Angle-Side) congruence criterion,the two triangles are congruent.
We match the vertices based on the given equalities:
- Vertex $A$ corresponds to vertex $P$
- Vertex $B$ corresponds to vertex $R$
- Vertex $C$ corresponds to vertex $Q$
Therefore,$\Delta ABC \cong \Delta PRQ$.

(C) Given that in $\Delta PQR$ and $\Delta XYZ$:
$1$. $\angle P = \angle Z$
$2$. $\angle Q = \angle Y$
$3$. $PQ = YZ$
By the $ASA$ (Angle-Side-Angle) congruence criterion,the correspondence must match the vertices in the same order.
Since $\angle P$ corresponds to $\angle Z$,$\angle Q$ corresponds to $\angle Y$,and the side $PQ$ corresponds to $ZY$ (or $YZ$),the remaining vertex $R$ must correspond to $X$.
Therefore,the correspondence is $P \leftrightarrow Z$,$Q \leftrightarrow Y$,and $R \leftrightarrow X$.
Thus,$\Delta PQR \cong \Delta ZYX$.

(D) Given that in $\Delta ABC$ and $\Delta XYZ$:
$1$. $\angle A = \angle X$
$2$. $\angle C = \angle Z$
$3$. $AB = XY$
By the Angle-Angle-Side $(AAS)$ congruence criterion,since two angles and the included side between them are not equal,we look at the correspondence.
Here,$\angle A$ corresponds to $\angle X$,$\angle C$ corresponds to $\angle Z$,and the side $AB$ corresponds to $XY$.
Since $\angle A = \angle X$ and $\angle C = \angle Z$,the third angle $\angle B$ must be equal to $\angle Y$ (by Angle Sum Property).
Thus,the correspondence is $A \leftrightarrow X$,$B \leftrightarrow Y$,and $C \leftrightarrow Z$.
Therefore,$\Delta ABC \cong \Delta XYZ$.

(A) Given that $\frac{AB}{PQ} = \frac{BC}{RQ} = \frac{AC}{PR} = 1$.
This implies $AB = PQ$,$BC = RQ$,and $AC = PR$.
By the $SSS$ (Side-Side-Side) congruence criterion,if three sides of one triangle are equal to the three corresponding sides of another triangle,then the triangles are congruent.
Matching the vertices based on the equality of sides:
$AB = PQ$ (Vertex $A$ corresponds to $P$,$B$ to $Q$)
$BC = RQ$ (Vertex $B$ corresponds to $R$,$C$ to $Q$)
$AC = PR$ (Vertex $A$ corresponds to $P$,$C$ to $R$)
Thus,$\Delta ABC \cong \Delta PQR$ is incorrect based on the ratio provided. Let's re-evaluate the correspondence: $A \leftrightarrow P$,$B \leftrightarrow Q$,$C \leftrightarrow R$ is not possible since $BC=RQ$ and $AC=PR$.
Actually,$AB=PQ$,$BC=RQ$,$AC=PR$ implies $\Delta ABC \cong \Delta PQR$ is not the correct order. Let's check: $A$ corresponds to $P$,$B$ corresponds to $Q$,$C$ corresponds to $R$. Wait,$BC=RQ$ means $B$ corresponds to $R$ and $C$ corresponds to $Q$. $AC=PR$ means $A$ corresponds to $P$ and $C$ corresponds to $R$. $AB=PQ$ means $A$ corresponds to $P$ and $B$ corresponds to $Q$.
Matching: $A \leftrightarrow P$,$B \leftrightarrow Q$,$C \leftrightarrow R$. Therefore,$\Delta ABC \cong \Delta PQR$.

(B) Given: In $\Delta ABC$ and $\Delta PQR$,$\angle B = \angle P = 90^{\circ}$.
$AC = RQ$ (Hypotenuse of $\Delta ABC$ = Hypotenuse of $\Delta PQR$).
$AB = RP$ (One side of $\Delta ABC$ = One side of $\Delta PQR$).
By the $RHS$ (Right angle-Hypotenuse-Side) congruence criterion,the triangles are congruent.
Matching the vertices: $A$ corresponds to $R$,$B$ corresponds to $P$,and $C$ corresponds to $Q$.
Therefore,$\Delta ABC \cong \Delta RPQ$.

(D) Given that $\Delta PRQ \cong \Delta ZXY$.
By the property of congruent triangles,the corresponding parts of congruent triangles are equal $(CPCT)$.
Comparing the vertices of the two triangles:
$P$ corresponds to $Z$
$R$ corresponds to $X$
$Q$ corresponds to $Y$
Therefore,$\angle P = \angle Z$,$\angle R = \angle X$,and $\angle Q = \angle Y$.
Since the question asks for the angle in $\Delta XYZ$ that is equal to $\angle R$,we look at the correspondence: $\angle R$ corresponds to $\angle X$.
Thus,$\angle R = \angle X$.