Question

AX and DY are altitudes of two similar triangles ΔABC and ΔDEF. Prove that AX : DY = AB : DE.

Answer 1

In the attachment I have answered this problem.   Concept:   If two triangles are similar their corresponding sides are proportional.   AAA similarity:   If each angle of one triangle is equal to each of angle of other then the triangles are similar.  See the attachment for detailed solution

Answer 2

Given, Δ ABC ∼ Δ DEF
⇒ ∠ ABC = ∠ DEF
⇒ consider Δ ABX and Δ DEY
⇒ ∠ ABX = ∠ DEY
⇒ ∠ AXB = ∠ DYE = 90°
⇒ ∠ BAX = ∠ EDY
⇒ By A-A-A property we have in two triangles if the angles are equal, then sides opposite to the equal angles are in the same ratio (or proportional) and hence the triangles
⇒ Δ ABX ∼ Δ DEY.
so, $\frac{AB}{DE}=\frac{BX}{EY}=\frac{AX}{DY}$
⇒ $\frac{AX}{DY}=\frac{AB}{DE}$
∴ AX:DY = AB:DE
Hence proved