The diagonals of a quadrilateral abcd intersect at o prove that

Transcript Ex 6.2, 10 The diagonals of a quadrilateral ABCD intersect each other at the point O such that 𝐴𝑂/𝐵𝑂 = 𝐶𝑂/𝐷𝑂 . Show that ABCD is a trapezium Given: ABCD is a quadrilateral where diagonals AC & BD intersect at O & 𝐴𝑂/𝐵𝑂=𝐶𝑂/𝐷𝑂 To prove: ABCD is a trapezium Construction: Let us draw a line EF II AB passing through point O. Proof: Given 𝐴𝑂/𝐵𝑂=𝐶𝑂/𝐷𝑂 ⇒ 𝐴𝑂/𝐶𝑂=𝐵𝑂/𝐷𝑂 Now, in ∆ 𝐴𝐷𝐵 EO II AB 𝐴𝐸/𝐷𝐸=𝐵𝑂/𝐷𝑂 ⇒ 𝐴𝐸/𝐷𝐸=𝐴𝑂/𝐶𝑂 Thus in Δ ADC, Line EO divides the triangle in the same ratio ∴ EO II DC Now, EO II DC But, we know that EO II AB ⇒ EO II AB II DC ⇒ AB II DC Hence, one pair of opposite sides of quadrilateral ABCD are parallel Therefore ABCD is a trapezium . Hence proved Show More

It is given that, A r e a ( Δ A O D ) = A r e a ( Δ B O C ) A r e a ( Δ A O D ) + A r e a ( Δ A O B ) = A r e a ( Δ B O C ) + A r e a ( Δ A O B ) [Adding Area ( Δ A O B ) to both sides] A r e a ( Δ A D B ) = A r e a ( Δ A C B ) We know that triangles on the same base having areas equal to each other lie between the same parallels. Therefore, these triangles, Δ A D B and Δ A C B, are lying between the same parallels. i.e., AB || CD Therefore, ABCD is a trapezium.

A quadrilateral $ABCD$ is formed from four distinct points (called the vertices), no three of which are collinear, and from the segments $AB$, $CB$, $CD$, and $DA$ (called the sides), which have no intersections except at those endpoints labeled by the same letter. The notation for this quadrilateral is not unique- e.g., quadrilateral $ABCD$ = quadrilateral $CBAD$. Two vertices that are endpoints of a side are called adjacent; otherwise the two vertices are called opposite. The remaining pair of segments $AC$ and $BD$ formed from the four points are called diagonals of the quadrilateral; they may or may not intersect at some fifth point. If $X$, $Y$, $Z$ are the vertices of quadrilateral $ABCD$ such that $Y$ is adjacent to both $X$ and $Z$, then angle $XYZ$ is called an angle of the quadrilateral; if $W$ is the fourth vertex, then angle $XWZ$ and angle $XYZ$ are called opposite angles. The quadrilaterals of main interest are the convex ones. By definition, they are the quadrilaterals such that each pair of opposite sides, e.g., $AB$ and $CD$, has the property that $CD$ is contained in one of the half-planes bounded by the line through $A$ and $B$, and $AB$ is contained in one of the half-planes bounded by the line through $C$ and $D$. a) Using Pasch's theorem, prove that if one pair of opposite sides has this property, then so does the other pair of opposite sides. b) Prove, using the crossbar theorem, that the following are equivalent: • The quadrilateral is convex. • Eac...