1. **Problem Statement:** We have a square ABCD with diagonals AC and BD intersecting at N. Points E, F, and G lie on sides AD, AC, and CD respectively such that DEFG is a rectangle. (a) Prove that $AE = EF$. (b) Prove that $\triangle AEN \cong \triangle DGN$. (c)(i) Given points H on AD and K on EN such that $HK \perp EN$ and $HK = EN$, determine if GHKN is a parallelogram. (c)(ii) Find the area of quadrilateral GHEN given $HG = 30\sqrt{2}$ and $EG = 70$. --- 2. **(a) Prove $AE = EF$** - Given angles at F: $\angle EAF = 45^\circ$, $\angle AEF = 90^\circ$, $\angle AFE = 45^\circ$. - Since $\angle EAF = \angle AFE = 45^\circ$, triangle AEF is isosceles with $AE = EF$ (sides opposite equal angles are equal). 3. **(b) Prove $\triangle AEN \cong \triangle DGN$** - Given: - $AE = DG$ (sides of rectangle DEFG) - $AN = DN$ (N is midpoint of diagonals in square) - $\angle EAN = \angle GDN$ (corresponding angles in square) - By SAS (Side-Angle-Side) congruence criterion, $\triangle AEN \cong \triangle DGN$. 4. **(c)(i) Is GHKN a parallelogram?** - Given: - $HK \perp EN$ - $HK = EN$ - From (b), $EN = GN$ - Since $HK = EN = GN$, and $HK \perp EN$, GHKN has two pairs of equal sides but one pair is perpendicular. - For GHKN to be a parallelogram, opposite sides must be parallel. - Here, $HK \perp EN$ implies $HK$ is not parallel to $GN$. - Therefore, GHKN is **not** a parallelogram. 5. **(c)(ii) Find area of quadrilateral GHEN** - Given: - $HG = 30\sqrt{2}$ - $EG = 70$ - Quadrilateral GHEN can be split into triangles or use vector/coordinate geometry. - Since $HG$ and $EG$ are adjacent sides of rectangle DEFG, area of GHEN equals area of rectangle DEFG minus triangle or can be computed as: - Area $= \frac{1}{2} \times HG \times EG = \frac{1}{2} \times 30\sqrt{2} \times 70 = 15 \times 70 \times \sqrt{2} = 1050\sqrt{2}$. --- **Final answers:** (a) $AE = EF$ (b) $\triangle AEN \cong \triangle DGN$ (c)(i) GHKN is not a parallelogram. (c)(ii) Area of GHEN is $1050\sqrt{2}$.