Similar Shapes Worksheets| Printable PDF Congruence and Similarity ... - Free Printable

Problem Overview:

The task involves two sections:
1. Section A: Prove that ΔABC is similar to ΔAXY for three given diagrams.
2. Section B: Calculate the required lengths using properties of similar triangles.

Let's solve each part step by step.

---

Section A: Prove that ΔABC is similar to ΔAXY

#### 1) Diagram 1:
- Given: Line XY is parallel to line BC.
- To Prove: ΔABC ~ ΔAXY.

Proof:
- Since XY || BC, by the Basic Proportionality Theorem (Thales' Theorem), we know that:
\[
\frac{AX}{AB} = \frac{AY}{AC}
\]
- Additionally, ∠BAC is common to both triangles ΔABC and ΔAXY.
- Therefore, by the AA (Angle-Angle) similarity criterion, ΔABC ~ ΔAXY.

Conclusion: ΔABC is similar to ΔAXY because they have two pairs of corresponding angles equal.

---

#### 2) Diagram 2:
- Given: Line XY intersects sides AB and AC at points X and Y, respectively, such that AX/AB = AY/AC.
- To Prove: ΔABC ~ ΔAXY.

Proof:
- We are given that:
\[
\frac{AX}{AB} = \frac{AY}{AC}
\]
- Additionally, ∠BAC is common to both triangles ΔABC and ΔAXY.
- Therefore, by the SAS (Side-Angle-Side) similarity criterion, ΔABC ~ ΔAXY.

Conclusion: ΔABC is similar to ΔAXY because they have a pair of corresponding angles equal and the sides including these angles are proportional.

---

#### 3) Diagram 3:
- Given: ΔABC with side lengths AB = 8 cm, AC = 10 cm, and BC = 6 cm. Points X and Y divide AB and AC in the ratio 7:1.
- To Prove: ΔABC ~ ΔAXY.

Proof:
- Points X and Y divide AB and AC in the ratio 7:1. This means:
\[
\frac{AX}{AB} = \frac{7}{8} \quad \text{and} \quad \frac{AY}{AC} = \frac{7}{10}
\]
- Since XY || BC (by the Basic Proportionality Theorem), the triangles ΔABC and ΔAXY are similar by the AA similarity criterion.

Conclusion: ΔABC is similar to ΔAXY because they have two pairs of corresponding angles equal.

---

Section B: Calculate the Required Lengths

#### 1) Diagram 1:
- Given: WXYZ is a parallelogram with WX = 6 cm, WZ = 14 cm, and XY = 9 cm.
- To Find: XZ and XV.

Solution:
- In a parallelogram, opposite sides are equal. Therefore:
\[
XY = WZ = 9 \, \text{cm} \quad \text{and} \quad WX = ZY = 6 \, \text{cm}
\]
- Since WXYZ is a parallelogram, the diagonals bisect each other. Let the diagonals intersect at point V. Then:
\[
XV = VZ \quad \text{and} \quad WV = VX
\]
- Using the property of the diagonals of a parallelogram:
\[
XZ = \sqrt{WX^2 + WZ^2 - 2 \cdot WX \cdot WZ \cdot \cos(\angle W)}
\]
However, since we are not given the angle, we use the fact that the diagonals bisect each other:
\[
XZ = \sqrt{WX^2 + WZ^2} = \sqrt{6^2 + 14^2} = \sqrt{36 + 196} = \sqrt{232} = 2\sqrt{58} \, \text{cm}
\]
And:
\[
XV = \frac{XZ}{2} = \frac{2\sqrt{58}}{2} = \sqrt{58} \, \text{cm}
\]

Conclusion:
\[
XZ = 2\sqrt{58} \, \text{cm}, \quad XV = \sqrt{58} \, \text{cm}
\]

#### 2) Diagram 2:
- Given: ΔPQR with PQ = 12 cm, PR = 6 cm, and QR = 9 cm. ST is parallel to QR.
- To Find: ST, RS, and RT.

Solution:
- Since ST || QR, by the Basic Proportionality Theorem, the triangles ΔPST and ΔPQR are similar.
- Let the ratio of similarity be \( k \). Then:
\[
\frac{PS}{PQ} = \frac{PT}{PR} = \frac{ST}{QR} = k
\]
- We need more information to find the exact lengths. However, if we assume the ratio is given or can be determined, we can use the similarity ratios to find the lengths.

Conclusion: Without additional information, we cannot determine the exact lengths of ST, RS, and RT. If more information is provided, we can solve for them using the similarity ratios.

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Final Answers:

\[
\boxed{
\begin{array}{l}
\text{Section A:} \\
1) \, \Delta ABC \sim \Delta AXY \, (\text{AA similarity}) \\
2) \, \Delta ABC \sim \Delta AXY \, (\text{SAS similarity}) \\
3) \, \Delta ABC \sim \Delta AXY \, (\text{AA similarity}) \\
\text{Section B:} \\
1) \, XZ = 2\sqrt{58} \, \text{cm}, \, XV = \sqrt{58} \, \text{cm} \\
2) \, \text{More information needed for ST, RS, and RT.}
\end{array}
}
\]