There is a triangle ABC. It has points X and Y on AB and CB so that XY is parallel to BC. Also, area of triangle BXY is two times the area of quad. ACYX. What is the ratio of AX to AB ?
karthik4297:
check your question when point is on AB then line never will be parallel to AB
Oops... thanks for pointing out
no worries its happens
question is wrong
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Question is wrongly written. I correct that and solve that. In a ΔABC, X is on AB and Y is on BC so that XY is parallel to AC. Area Quadrilateral ACXY = 2 times area Δ BXY. Now what is AX / AB?
Answer: Let BX = t AB. Angle BXY = angle A and angle BYX = angle C. So BXY and BAC are similar triangles. THe sides are proportional. So BY = t BC and XY = t AC. Now find semi-perimeter s" of ΔBXY = t(AB+BC+AC) = t s.
Area of ΔBXY = √s" (s"-BX) (s"-BY) (s"-XY) = √t s t(s-BA) t(s-BC) t(s-AC)
= t² area ΔABC
so t² = 1/2 ==> t =1/√2 so BX = 1/√2 BA
so AX = BA - BX = BA * (√2 - 1)/√2
Answer: Let BX = t AB. Angle BXY = angle A and angle BYX = angle C. So BXY and BAC are similar triangles. THe sides are proportional. So BY = t BC and XY = t AC. Now find semi-perimeter s" of ΔBXY = t(AB+BC+AC) = t s.
Area of ΔBXY = √s" (s"-BX) (s"-BY) (s"-XY) = √t s t(s-BA) t(s-BC) t(s-AC)
= t² area ΔABC
so t² = 1/2 ==> t =1/√2 so BX = 1/√2 BA
so AX = BA - BX = BA * (√2 - 1)/√2
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