Question

in triangle ABC the line segment XY is parallel to side AC of triangle ABC and it divides the triangle into two parts of equal areas.find the ratio of AX/AB

Answer 1

Step-by-step explanation:

∵ XY || AC

∠BXY = ∠BAC (corresponding angles)

∠BYX = ∠BCA (,,)

∠XBY = ∠ABC (same angle)

∴ ΔBXY and ΔBAC are congruent

Again, Area.BXY = Area.XACY -----(1)

And, Area.BAC = Area.BXY + Area.XACY

= Area.BXY + Area. BXY (From 1)

= 2× Area.BXY

∴ AX/AB = (AB-BX)/AB

= 1-(BX/AB)

= 1-(BX/2BX)

= 1-1/2

= 1/2