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In $\triangle ABC$ and $\triangle DEF$, it is being given that: $AB = 5\ cm, BC = 4\ cm$ and $CA = 4.2\ cm; DE = 10\ cm, EF = 8\ cm$ and $FD = 8.4\ cm$. If $AL \perp BC$ and $DM \perp EF$, find $AL : DM$.
Given:
$AB = 5\ cm, BC = 4\ cm$ and $CA = 4.2\ cm; DE = 10\ cm, EF = 8\ cm$ and $FD = 8.4\ cm$.
$AL \perp BC$ and $DM \perp EF$.
To do:
We have to find $AL : DM$.
Solution:
In $\triangle ABC$ and $\triangle DEF$,
$\frac{AB}{DE}=\frac{5}{10}=\frac{1}{2}$
$\frac{AC}{DF}=\frac{4.2}{8.4}=\frac{1}{2}$
$\frac{BC}{EF}=\frac{4}{8}=\frac{1}{2}$
$\frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}$
Therefore,
By SSS similarity,
$\triangle ABC \sim\ \triangle DEF$
This implies,
$\angle C=\angle F$
In $\triangle ALC$ and $\triangle DMF$,
$\angle ALC=\angle DMF=90^o$
$\angle C=\angle F$
Therefore,
$\triangle ALC \sim\ \triangle DMF$
This implies,
$\frac{AC}{DF}=\frac{AL}{DM}$
$\frac{AL}{DM}=\frac{1}{2}$
$AL:DM=1:2$.
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