Question

The altitude $BN$ and $CM$ of $△ABC$ meet at $H$. Prove that
(i) $CN\times HM=BM\times HN$
(ii) $HC/HB=\sqrt{\left[\right(CN\times HN\left)/\right(BM\times HM\left)\right]}$
(iii) $△MHN\sim △BHC$.

Answer 1

Consider the $△ABC$,
Where, the altitude $BN$ and $CM$ of $△ABC$ meet at $H$.
and construction: join $MN$
(i) We have to prove that, $CN\times HM=BM\times HN$
In $△BHM$ and $△CHN$
$\angle BHM=\angle CHN$ [because vertically opposite angles are equal]
$\angle M=\angle N$ [both angles are equal to ${90}^{\circ }$]
Therefore, $△BHM\sim △CHN$
So, $HM/HN=BM/CN=HB/HC$
Then, by cross multiplication we get
$CN\times HM=BM\times HN$
(ii) Now, $HC/HB=\sqrt{(HN\times CN)/(HM\times BM)}$
$=\sqrt{(CN\times HN)/(BM\times HM)}$
Because, $M$ and $N$ divide $AB$ and $AC$ in the same ratio.
(iii) Now consider $△MHN$ and $△BHC$
$\angle MHN=\angle BHC$ [because vertically opposite angles are equal]
$\angle MNH=\angle HBC$ [because alternate angles are equal]
Therefore, $△MHN\sim △BHC$.