Question

AX and DYare altitudesof two similar triangles ABC and DEF. Prove that AX:DY=AB:DC

Answer 1

GIVEN :

AX and DY are the altitudes of ΔABC and ΔDEF. ΔABC~ ΔDEF.

TO PROVE :

AX : DY = AB : DE

PROOF :

Since ΔABC~ ΔDEF

∠A = ∠D ; ∠B = ∠E ; ∠C = ∠F     .......(1)

In ΔAXB and ΔDYE

∠B = ∠E                   [using(1)]

∠AXB = ∠DYE [90° each]

so, ΔAXB~ ΔDYE   [AA similarity]

⇒AX/DY = XB/YE = AB/DE (Corresponding sides of 2 similar Δ's are proportional)

⇒AX/DY  = AB/DE

⇒AX : DY = AB : DE  PROVED

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Answer 2

AX and DY are the altitudes of ΔABC and ΔDEF. ΔABC~ ΔDEF.

AX : DY = AB : DE

Since ΔABC~ ΔDEF

∠A = ∠D ; ∠B = ∠E ; ∠C = ∠F     .......(1)

In ΔAXB and ΔDYE

∠B = ∠E                   [using(1)]

∠AXB = ∠DYE [90° each]

so, ΔAXB~ ΔDYE   [AA similarity]

AX/DY = XB/YE = AB/DE (Corresponding sides of two similar Δ's are proportional)

AX/DY  = AB/DE

AX : DY = AB : DE