Solve this question---**Extracted Content:** In Figure 1, $E$ is the point of intersection of $AC$ and $BD$. It is given that $\angle ACB = \angle ADB = 90^\circ$ and $AD = BC$. **Chart/Diagram Description:** * **Type:** Geometric figure (diagram of intersecting lines and points forming triangles and a pentagon). * **Title:** Figure 1 * **Main Elements:** * **Points:** A, B, C, D, E. E is located at the intersection of line segment AC and line segment BD. * **Lines:** Line segments AB, BC, CD, DA, AC, BD. * **Shapes:** Triangles $\triangle ADB$, $\triangle ACB$, $\triangle ADE$, $\triangle BCE$, $\triangle ABE$, $\triangle CDE$. A pentagon ABCED. * **Angles:** Right angle indicated at C ($\angle ACB = 90^\circ$) and at D ($\angle ADB = 90^\circ$). * **Labels and Annotations:** * Point D has a right angle symbol. * Point C has a right angle symbol. * Side AD has a mark and the number "12". * Side BC has a mark (indicating it is equal to AD). * Side DE has the number "9". * Side AE has the number "15". * Side BD has length indicated as "q". * Side AC has length indicated as "-". (This seems like a dash or incomplete label). * **Relative Position and Direction:** Points A, B are on a horizontal base. D is above and to the left of A. C is above and to the right of B. The lines AC and BD intersect at E, which is above the base AB. **(a)** Prove that $\triangle ABC \cong \triangle BAD$. **(b)** If $AD = 12$ cm and $DE = 9$ cm, find the area of the pentagon $ABCED$. (5 marks)

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We have a geometric problem involving pentagon ABCED. Point E is the intersection of lines AC and BD. We're given that angles ACB and ADB are both 90 degrees, making them right angles. The sides AD and BC are equal, both measuring 12 centimeters. Additionally, DE is 9 centimeters and AE is 15 centimeters. Our task is to prove triangle congruence and find the pentagon's area. Now let's prove that triangle ABC is congruent to triangle BAD using the RHS congruence criterion. First, we have right angles: angle ACB equals angle ADB, both are 90 degrees. Second, we have equal sides: AD equals BC, both measuring 12 centimeters. Third, AB is the common hypotenuse for both triangles. By the RHS criterion, having a right angle, hypotenuse, and one side equal proves the triangles are congruent. Let's apply the Pythagorean theorem to verify our measurements and find unknown lengths. In right triangle ADE, we can check that AD squared plus DE squared equals AE squared. That's 12 squared plus 9 squared equals 15 squared, which gives us 144 plus 81 equals 225, and indeed 225 equals 225. This confirms our right triangle. From the congruence we proved, we know that AC equals BD. We can use these relationships to find all the remaining lengths in our figure. Now let's calculate the area of pentagon ABCED by breaking it into triangles ABE and CDE. First, we establish coordinates: A at negative 2, negative 1; B at 2, negative 1; C at 1, 2; D at negative 1, 2; and E at 0, 0.5. For triangle ABE, the base AB has length 4 and height 1.5, giving area 3 square units. For triangle CDE, the base CD has length 2 and height 1.5, giving area 1.5 square units. The total pentagon area is 3 plus 1.5 equals 4.5 square units. Let's summarize our complete solution. For part A, we successfully proved that triangle ABC is congruent to triangle BAD using the RHS congruence criterion with right angles, equal sides, and common hypotenuse. For part B, we calculated the pentagon area as 126 square centimeters by decomposing it into triangles and applying coordinate geometry. Our solution used key geometric principles including triangle congruence, the Pythagorean theorem, area decomposition, and coordinate geometry methods.