In the following figure, in \(\Delta ABC\) ray BD bisects \(\angle ABC\), A – D – C, seg ED \(\parallel\) side BC, A – E – B, then to prove that \(\dfrac{AB}{BC} = \dfrac{AE}{EB}\), complete the following activity.[Figure: \(\Delta ABC\) with A at th…

Q.3 (A)(i) In the following figure, in \(\Delta ABC\) ray BD bisects \(\angle ABC\), A – D – C, seg ED \(\parallel\) side BC, A – E – B, then to prove that \(\dfrac{AB}{BC} = \dfrac{AE}{EB}\), complete the following activity.

[Figure: \(\Delta ABC\) with A at the top, B at the lower left and C at the lower right; D is on side AC and ray BD is drawn and extended beyond D with an arrow; E is on side AB; seg ED is drawn parallel to BC; the two equal parts of \(\angle ABC\) are marked at B.]

Activity : In \(\Delta ABC\), ray BD bisects \(\angle B\).

\(\therefore \dfrac{AB}{BC} = \dfrac{\text{____}}{\text{____}}\) .............. (I) (Angle bisector theorem)

In \(\Delta ABC\), seg ED \(\parallel\) side BC.

\(\therefore \dfrac{AE}{EB} = \dfrac{\text{____}}{DC}\) .............. (II) ____

\(\therefore \dfrac{AB}{\text{____}} = \dfrac{\text{____}}{EB}\) .............. from (I) and (II)