Proving Triangle

Question

I spend so much time for proving this triangle and i still don't know.

Question :

Given Triangle ABC, AD and BE are altitudes of the triangle. Prove that Triangle DEC similarity with triangle ABC

Answer 1

Since $AD=AC\cdot cos C$ and $BE=BC\cdot cos C$ we have that $$\frac{AD}{BE}=\frac{AC}{BE}$$ And since the angle that form those two pairs are equal ($\angle C$) the result follows

Answer 2

Construct a figure from the data you have given, now follow these steps(you just have to prove equality of any two angles of respective triangles):

Since BE and AD are perpendiculars so you get ∠BEA = ∠ADB =90 Degrees.Now you can see that ABDE is a cyclic quadrilateral {as ∠BEA = ∠ADB(you can call them angles in the same segment)}

Now as you know sum of opposite angles in a cyclic quadrilateral is 180 degrees so ∠EDB= 180-∠A = ∠EDC. Therefore:

For triangles CDE and CAB: ∠CDE=∠A and ∠C = ∠C

Hence,triangles CDE and CAB are similar.