2. Two mark example problems I

1. \(D\) and \(E\) are respectively the points on the sides \(AB\) and \(AC\) of a \(\Delta ABC\) such that \(AB = 5.6 \ cm\), \(AD = 1.4 \ cm\), \(AC = 7.2 \ cm\) and \(AE = 1.8 \ cm\), show that \(DE || BC\).

 

Answer:

 

 

 

$\frac{\mathit{AD}}{\mathit{BD}}$ =≠ $\frac{\mathit{AE}}{\mathit{CE}}$

 

By Basic proportionalityConverse of basic proportionalityConverse of angle bisectorAngle bisector theorem, \(DE || BC\).

 

[Note: Enter the simplified ratio.]

 

 

2.

 

In above figure, \(DE || AC\) and \(DC || AP\). Prove that $\frac{\mathit{BE}}{\mathit{EC}}=\frac{\mathit{BC}}{\mathit{CP}}$.

Answer:

 

In \(\Delta BPA\), we have $\mathit{DC}\parallel i$.

 

By Basic proportionalityPythagoras theoremConverse of Basic proportionalityAngle bisector theorem:

 

$\frac{\mathit{BC}}{i}=\frac{i}{\mathit{DA}}$ - - - - - - (I)