Question

The altitudes BN and CM of $\Delta ABC$ meet at H. Prove that

(1) CN.HM=BM.HN

(2)$\frac{HC}{HB}=\sqrt{\frac{CN.HN}{BM.HM}}$

(3)$\Delta MHN\sim \Delta BHC$

Answer 1

given, $\left(BN\perp AC\right)$ & $\left(CM\perp AB\right)$

$\left(A\right)$ To prove $\left(CN\right)\left(HM\right)=\left(BM\right)\left(HN\right)$

Proof In $BHM$ & $CHN$

$\angle M=\angle N={90}^o$

$\angle BHM=\angle CHN....$ (velocity opposite angles)

$\therefore △BHM\sim △CHN$ by $A-A$centre

$\left(B\right)$ Hence, $\frac{BM}{CN},\frac{HM}{HN}....\left(i\right)\Rightarrow \left(CN\right)\left(HM\right)=\left(BM\right)\left(HN\right)$

$\Rightarrow \frac{BM}{HM}=\frac{CN}{NH}$

$\Rightarrow \frac{BM+HM}{HM}=\frac{CN+NH}{NH}.....$ (Applying components)

$\Rightarrow \left(\frac{GH}{HM}\right)\left(\frac{CH}{NH}\right)$

$\Rightarrow \left(\frac{BH}{CH}\right)=\left(\frac{MH}{NH}\right)...\left(ii\right)$

$\Rightarrow \frac{CH}{BH}=\frac{NH}{MH}$

$\Rightarrow \frac{CH}{BH}=\sqrt{\left(\frac{NH}{MH}\right)\left(\frac{NH}{MH}\right)}$

$\Rightarrow \left(\frac{CH}{BH}\right)=\sqrt{\frac{\left(NH\right)\left(CN\right)}{\left(MN\right)\left(BM\right)}}.....$ [from equation $i$]

$\left(C\right)$ To prove, $△MHN\sim △BHC$

Proof In $△MHN$ & $△BHC$

$\Rightarrow \angle MHN=\angle BHC$ (vertically opposite angles)

& $\left(\frac{BH}{CH}\right)=\left(\frac{MH}{NH}\right)$.... [from $\left(ii\right)$]

$\therefore △MHN\sim △BHC$ by $SAS$ written