Question
In $\triangle \mathrm{ABC}$, ray $\mathrm{BD}$ bisects $\angle \mathrm{ABC}$. $A-D-C$, side DE $\|$ side $B C, A-E-B$ then prove that, $\frac{A B}{B C}=\frac{A E}{E B}$
✓
Answer
In $\triangle A B C$, ray $B D$ bisects $\angle A B C$.
Points $A-D-C$ are collinear, $D E \| B C$, and points $A-E-B$ are collinear.
To Prove : $\frac{A B}{B C}=\frac{A E}{E B}$
Proof:
In $\triangle A B C$, ray $B D$ bisects $\angle A B C$.
$\therefore \frac{A B}{B C}=\frac{A D}{D C} \quad \ldots \ldots$
(I) (Angle bisector theorem)
In $\triangle A B C$, since $D E \| B C$,
$\frac{A E}{E B}=\frac{A D}{D C} \quad \ldots \ldots$
(II) (Basic proportionality theorem)
From equations (I) and (II),
$\frac{A B}{B C}=\frac{A E}{E B}$
Points $A-D-C$ are collinear, $D E \| B C$, and points $A-E-B$ are collinear.
To Prove : $\frac{A B}{B C}=\frac{A E}{E B}$
Proof:
In $\triangle A B C$, ray $B D$ bisects $\angle A B C$.
$\therefore \frac{A B}{B C}=\frac{A D}{D C} \quad \ldots \ldots$
(I) (Angle bisector theorem)
In $\triangle A B C$, since $D E \| B C$,
$\frac{A E}{E B}=\frac{A D}{D C} \quad \ldots \ldots$
(II) (Basic proportionality theorem)
From equations (I) and (II),
$\frac{A B}{B C}=\frac{A E}{E B}$
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