Asked in: March 2024 and June 2026 board papers In Δ ABC, ray BD bisects ∠ ABC, A – D – C, seg DE || side BC, A – E – B. Then, to show AB/BC = AE/EB, complete the following activity. [Figure: Δ ABC with D on AC so that BD bisects ∠ B, and E on AB so that DE is parallel to BC.] Activity: In Δ ABC, ray BD bisects ∠ B. therefore ____/BC = AD/DC ...... (I) (____). In Δ ABC, DE || BC therefore ____/EB = AD/DC ...... (II) (____). therefore AB/____ = ____/EB [from (I) and (II)]

Answer: AB/BC = AE/EB (blanks: AB, angle bisector theorem; AE, basic proportionality theorem; BC, AE) In Δ ABC, ray BD bisects ∠ B. therefore dfracboxedABBC = AD/DC ...... (I) (theorem of angle bisector of a triangle) In Δ ABC, seg DE || side BC. therefore dfracboxedAEEB = AD/DC ...... (II) (basic proportionality theorem) From (I) and (II), both ratios equal AD/DC, so dfracABboxedBC = dfracboxedAEEB.

Chapter: Similarity· Mathematics Part 2 (Geometry)

Asked in: March 2024 and June 2026 board papersIn \(\Delta ABC\), ray BD bisects \(\angle ABC\), A – D – C, seg DE || side BC, A – E – B. Then, to show \(\dfrac{AB}{BC} = \dfrac{AE}{EB}\), complete the following activity.[Figure: \(\Delta ABC\) with…

Asked in: March 2024 and June 2026 board papers

In \(\Delta ABC\), ray BD bisects \(\angle ABC\), A – D – C, seg DE || side BC, A – E – B. Then, to show \(\dfrac{AB}{BC} = \dfrac{AE}{EB}\), complete the following activity.

[Figure: \(\Delta ABC\) with D on AC so that BD bisects \(\angle B\), and E on AB so that DE is parallel to BC.]

Activity: In \(\Delta ABC\), ray BD bisects \(\angle B\). \(\therefore \dfrac{\text{____}}{BC} = \dfrac{AD}{DC}\) ...... (I) (____). In \(\Delta ABC\), DE || BC \(\therefore \dfrac{\text{____}}{EB} = \dfrac{AD}{DC}\) ...... (II) (____). \(\therefore \dfrac{AB}{\text{____}} = \dfrac{\text{____}}{EB}\) [from (I) and (II)]

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Answer: \(\dfrac{AB}{BC} = \dfrac{AE}{EB}\) (blanks: AB, angle bisector theorem; AE, basic proportionality theorem; BC, AE)

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

\(\therefore \dfrac{\boxed{AB}}{BC} = \dfrac{AD}{DC}\) ...... (I) (theorem of angle bisector of a triangle)

In \(\Delta ABC\), seg DE || side BC.

\(\therefore \dfrac{\boxed{AE}}{EB} = \dfrac{AD}{DC}\) ...... (II) (basic proportionality theorem)

From (I) and (II), both ratios equal \(\dfrac{AD}{DC}\), so \(\dfrac{AB}{\boxed{BC}} = \dfrac{\boxed{AE}}{EB}\).

Doubtly AI · 29 Oct 2026 AI-verified · Nov 2026 Report

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Asked in: March 2024 and June 2026 board papersIn \(\Delta ABC\), ray BD bisects \(\angle ABC\), A – D – C, seg DE || side BC, A – E – B. Then, to show \(\dfrac{AB}{BC} = \dfrac{AE}{EB}\), complete the following activity.[Figure: \(\Delta ABC\) with…

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