Answer:
By the given figure,
In triangles ABC and CDE
m ∠ABC = m∠CDE = 90°
⇒ ∠ABC ≅ ∠CDE
Also, BC ║ DE ( given )
By the corresponding angles theorem,
∠ACB ≅ ∠CED
Thus, by AA similarity postulate,
,
By the similarity,
![(AB)/(BC)=(BC)/(DE)=(CA)/(EC)](https://img.qammunity.org/2020/formulas/mathematics/high-school/6a3uf1qo833z5hxb1y4ke3veg5qq2y1bvy.png)
3.
,
4. AA similarity postulate,
5. Congruent angles are,
∠ABC ≅ ∠CDE,
∠ACB ≅ ∠CED