NCERT Example 2 — Triangles

NCERT Example

Example 2

basic proportionality theorem · trapezium · parallel lines

ABCD is a trapezium with AB ∥ DC. E and F are points on the non-parallel sides AD and BC respectively such that EF is parallel to AB. Show that AE/ED = BF/FC.

Trapezium ABCD with AB parallel to DC and segment EF parallel to both bases. — Figure 6-14 from the supplied NCERT chapter.

Solution

Given:

ABCD is a trapezium with AB parallel to DC; E and F lie on AD and BC, and EF is parallel to AB.

To prove:

AE/ED = BF/FC.

Construction:

Join A to C. Mark G where AC meets EF.

The same trapezium ABCD with diagonal AC meeting EF at G. — Figure 6-15 from the supplied NCERT chapter.

Proof:

Since AB ∥ DC and EF ∥ AB, we also have EF ∥ DC.

In △ADC, EG ∥ DC. By BPT:

\frac{AE}{ED}=\frac{AG}{GC}\qquad ...(1)

In △CAB, GF ∥ AB. By BPT:

\frac{CG}{GA}=\frac{CF}{FB}

Taking reciprocals:

\frac{AG}{GC}=\frac{BF}{FC}\qquad ...(2)

From (1) and (2):

\frac{AE}{ED}=\frac{BF}{FC}

Final answer

Therefore, AE/ED = BF/FC.

Example 2

Example 2

Solution

Final answer

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