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P-2 Similarity — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsP-2 Similarity2 Marks
Question
In $\triangle A B C$, ray $B D$ bisects $\angle A B C . A-D-C$, side $D E \|$ side $B C, A-E-B$, then prove that $\frac{A B}{B C}=\frac{A E}{E B}$
Image
In $\triangle A B C$, ray $B D$ bisects $\angle B$.
$\therefore \quad \frac{ AB }{ BC }=\frac{ AD }{ DC }$
In $\triangle ABC , DE \| BC$
$\therefore \quad \frac{ AE }{ EB }=\frac{ AD }{ DC }$
$\therefore \frac{ AB }{ ⬜ }=\frac{ ⬜ }{ EB }$
[Given]
[Angle biscetor theorem]
[Given]
⬜
[From (i) and (ii)]
✓

Answer

In $\triangle A B C$, ray $B D$ bisects $\angle A B C . A-D-C$, side $D E \|$ side $B C, A-E-B$, then prove that $\frac{A B}{B C}=\frac{A E}{E B}$
Image
In $\triangle A B C$, ray $B D$ bisects $\angle B$.
$\therefore \quad \frac{ AB }{ BC }=\frac{ AD }{ DC }$
In $\triangle ABC , DE \| BC$
$\therefore \quad \frac{ AE }{ EB }=\frac{ AD }{ DC }$
$\therefore \frac{ AB }{ BC }=\frac{ AE }{ EB }$
[Given]
[Angle biscetor theorem]
[Given]
[Basic proportionality theorem]
[From (i) and (ii)]

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