1 In the given figure, line segment XY is parallel to side AC of triangle ABC and divides the - Maths - Triangles

Question

1 In the given figure, line segment XY is parallel to side AC of
triangle ABC and divides the triangle in two equal parts Find ratio
AX/AB

Shubham Chourasiya · Accepted Answer

Here in Î”XBY AND Î”ABC, âˆ B =âˆ B (Common in the both triangles) âˆ XYB =âˆ ACB (Corresponding angles) â‡’ Î”XBY âˆ¼ Î”ABC (angle angle similarity property) $\frac{a r (A B C)}{a r (X B Y)} = {(\frac{A B}{X B})}^{2}$ ------(1) (The ratio of the areas of 2 similar triangles is equal to the square of the ratio of their corresponding sides) Area of Î”ABC = 2 (Î”XBY ) (As XY divides the triangle in two equal parts) $\frac{a r (A B C)}{a r (X B Y)} = \frac{2}{1} (2) t h e r e f o r e f r o m (1) a n d (2), {(\frac{A B}{X B})}^{2} = \frac{2}{1} \Rightarrow \frac{A B}{X B} = \frac{\sqrt{2}}{1} \frac{X B}{A B} = \frac{1}{\sqrt{2}} \Rightarrow 1 - \frac{X B}{A B} = 1 - \frac{1}{\sqrt{2}} \frac{A B - X B}{A B} = \frac{\sqrt{2} - 1}{\sqrt{2}} \Rightarrow \frac{A X}{A B} = \frac{\sqrt{2} - 1}{\sqrt{2}} = \frac{2 - \sqrt{2}}{2}$

1. In the given figure, line segment XY is parallel to side AC of triangle ABC and divides the triangle in two equal parts.Find ratio AX/AB.