Question

In $△ABC$, $AL,BM$ and $CN$ are the altitudes which intersect at 'O'.
Prove that (i) $△AMB\sim △ANC$ (ii) $\frac{AN}{BN}.\frac{BL}{CL}.\frac{CM}{AM}=1$

Answer 1

In $△AMB$ and $△ANC$
$\angle AMB=\angle ANC={90}^o$ .......... ($\because$ Data)
$\angle BAM=\angle NAC$

$\angle ABM=\angle ACN$

$\therefore △AMB\sim △ANC$ ......... [By AAA rule]

Also, $△BLA\sim △BNC$
and $△BMC\sim △ALC$
Now, $△AMB\sim △ANC$

$\therefore \frac{AN}{AM}=\frac{AC}{AB}$

$△BLA\sim △BNC$

$\therefore \frac{BL}{BN}=\frac{AB}{BC}$

$△BMC\sim △ALC$

$\therefore \frac{CM}{CL}=\frac{BC}{AC}$

$\frac{AN}{AM}\times \frac{BL}{BN}\times \frac{CM}{CL}=\frac{AC}{AB}\times \frac{AB}{BC}\times \frac{BC}{AC}$

$\therefore \frac{AN}{BN}\times \frac{BL}{LC}\times \frac{CM}{AM}=1$

or $AN\times BL\times CM=BN\times LC\times AM$