NCERT Solutions
Class 10 - Mathematics - Chapter 6: TRIANGLES - Exercise 6.2

Question 9

Goal: Prove that the point of intersection of the diagonals of a trapezium (with one pair of parallel sides) divides the two diagonals in the same ratio, i.e. \(\dfrac{AO}{BO} = \dfrac{CO}{DO}\).

The hint suggests a construction, but the result can be proved neatly using coordinate geometry, keeping the geometry of a trapezium intact.

Step 1: Place the trapezium on the coordinate plane

Since AB \(\parallel\) DC, we can place the trapezium so that these bases are horizontal.

• Let \(A(0, 0)\) and \(B(b, 0)\), for some \(b > 0\).
• Let \(D(0, h)\) and \(C(c, h)\), for some \(h > 0\) and any real \(c\) (\(c \neq b\) in general).

Thus, AB is the segment from \((0,0)\) to \((b,0)\) and DC is the segment from \((0,h)\) to \((c,h)\); clearly AB \(\parallel\) DC.

Step 2: Find the coordinates of O, the intersection of diagonals

Diagonal AC joins \(A(0,0)\) and \(C(c,h)\). A general point on AC can be written in parametric form as:

\[ (x, y) = (tc,\; th), \quad 0 < t < 1. \]

Diagonal BD joins \(B(b,0)\) and \(D(0,h)\). A general point on BD can be written as:

\[ (x, y) = (b(1-s),\; hs), \quad 0 < s < 1. \]

At the intersection point O, these coordinates must be equal, so:

\[tc = b(1 - s), \quad th = hs.\]

Since \(h \neq 0\), from \(th = hs\) we get

\[t = s.\]

Thus, the same parameter (say \(t\)) describes O on both diagonals:

\[O(tc,\; th).\]

Step 3: Express the ratios along each diagonal

On diagonal AC, parameter \(t\) tells us how O divides AC:

• From A to O is a fraction \(t\) of AC, and from O to C is a fraction \(1 - t\).

Therefore,

\[\frac{AO}{OC} = \frac{t}{1 - t}. \quad (1)\]

On diagonal BD, parameter \(t\) again measures the fraction from B to D:

• From B to O is a fraction \(t\) of BD.
• From O to D is a fraction \(1 - t\) of BD.

Hence,

\[\frac{BO}{OD} = \frac{t}{1 - t}. \quad (2)\]

Step 4: Compare the ratios and derive the required result

From (1) and (2),

\[\frac{AO}{OC} = \frac{BO}{OD}.\]

This implies

\[AO \cdot OD = BO \cdot OC.\]

Divide both sides by \(BO \cdot OD\) (non-zero since O lies strictly between the endpoints on each diagonal):

\[\frac{AO}{BO} = \frac{OC}{OD}.\]

Thus, we have shown

\[\boxed{\dfrac{AO}{BO} = \dfrac{CO}{DO}}.\]

Conclusion: In a trapezium with one pair of parallel sides, the diagonals intersect each other in such a way that they are divided in the same ratio, as required.