Question

In ∆ABC, X is the middle point of AB. If XY ∥ BC, then prove that Y is the middle point of AC.
explain with steps
thank you​

Answer 1

$\large\underline{\sf{Given- }}$

➣ A ∆ABC such that X is the mid point of AB

and

➣ XY || BC

$\large\underline{\sf{To\:prove - }}$

➣ Y is the middle point of AC.

$\large\underline{\sf{Solution-}}$

Given that,

In ∆ABC

➣ X is the middle point of AB.

↝  It implies, AX = XB

Also, XY || BC

We know,

Basic Proportionality Theorem,

If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.

So, using this theorem, we get

$\rm :\longmapsto\:\dfrac{AX}{XB} = \dfrac{AY}{YC}$

$\rm :\longmapsto\:\dfrac{AX}{AX} = \dfrac{AY}{YC}$

$\red{\bigg \{ \because \: AX = XB\bigg \}}$

$\rm :\longmapsto\:1 = \dfrac{AY}{YC}$

$\rm :\longmapsto\:AY = YC$

$\bf\implies \:Y \: is \: midpoint \: of \: AC$

Hence, Proved

Additional Information :-

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.