(Q29) AX and DY are altitudes of two similar triangles ∆ABC and ∆DEF. Prove that AX : DY = AB : DE.

Solution :

ABCDEFXY

Given :∆ABC ∼ ∆DEF. and AX ⊥ BC and DY ⊥ EF.

R.T.P : AX : DY = AB : DE.

Proof :In ∆ABX and ∆DEY , ∠B = ∠

[∵ Corresponding angles of ∆ABC and ∆DEF]

∠AXB = ∠DYE [given]

∴ ∆ABX ∼ ∆DEY

(by A.A. similarity condition)

Hence,
DE
 =
BX
 =
DY

[∵ Ratios of corresponding sides of similar triangles are equal]

⇒ AX : DY = AB : DE