Question

In triangle ABC XY parallel to AC and XY divides the triangle into two equal parts then find the ratio of AX/XB

Answer 1

this is your answer......


In ΔABC, XY || AC and area of ΔBXY = area of quadrilateral XYCA ⇒ ar (ΔABC) = 2.ar (ΔBXY) ----------------(1) XY || AC and BA is a transversal. ⇒ ∠BXY = ∠BAC --------------------------- (2) So, In ΔBAC and ΔBXY, ∠XBY = ∠ABC (common angle) ∠BXY = ∠BAC [from equation (2)] ⇒ ΔBAC ~ ΔBXY ⇒ ar(ΔBAC) / ar(ΔBXY) = BA2 / BX2 ⇒ BA = √2 BX ⇒ BA = √2 (BA – AX) ⇒ (√2 – 1) BA = √2 AX ⇒ AX/XB = (√2 – 1) / √2


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