1. **Problem statement:** Given an isosceles triangle $\triangle ABC$ with $AB = AC$. Points $D$ and $E$ are taken on the rays opposite to $BC$ and $CB$ respectively such that $BD = CE$. Lines $BH \perp AD$ at $H$ and $CK \perp AE$ at $K$. Prove: 2. **Part a) Prove $\triangle ADE$ is isosceles.** - Since $AB = AC$ and $BD = CE$, triangles $\triangle ABD$ and $\triangle ACE$ are congruent by the Side-Side-Side (SSS) criterion. - Therefore, $AD = AE$. - Hence, $\triangle ADE$ is isosceles with $AD = AE$. 3. **Part b) Prove $\triangle BHD = \triangle CKE$.** - Since $BH \perp AD$ and $CK \perp AE$, angles $\angle BHD$ and $\angle CKE$ are right angles. - From part a), $AD = AE$. - Also, $BD = CE$ by given. - Triangles $\triangle BHD$ and $\triangle CKE$ have: - $BD = CE$ - $\angle BHD = \angle CKE = 90^\circ$ - $AD = AE$ - By the Hypotenuse-Leg (HL) theorem, $\triangle BHD \cong \triangle CKE$. 4. **Part c) Let $I$ be the intersection of $BH$ and $CK$. Prove $AI$ is perpendicular to $DE$ at the midpoint of $DE$.** - Since $\triangle BHD \cong \triangle CKE$, segments $BH$ and $CK$ intersect at $I$ such that $I$ lies on the line segment joining $B$ and $C$. - By symmetry and congruence, $I$ is the midpoint of segment $HK$. - Because $BH \perp AD$ and $CK \perp AE$, $AI$ is the altitude from $A$ to $DE$. - Therefore, $AI$ is perpendicular to $DE$ and passes through its midpoint. **Final answers:** - a) $\triangle ADE$ is isosceles with $AD = AE$. - b) $\triangle BHD \cong \triangle CKE$. - c) $AI \perp DE$ at the midpoint of $DE$.