# 01 Minimum distance between projection points on the legs of right triangle

**Problem**

From the right triangle ABC shown below, AB = 40 cm and BC = 30 cm. Points E and F are projections of point D from hypotenuse AC to the perpendicular legs AB and BC, respectively. How far is D from AB so that length EF is minimal?

**Solution**

EF and BD are diagonals of rectangle EBFD, thus, EF = BD. Length of EF is minimal only if BD is perpendicular to AC (BD as altitude through B of triangle ABC). Therefore,

$\sin \alpha = \dfrac{BD}{AB} = \dfrac{BC}{AC}$

$\sin \alpha = \dfrac{BD}{AB} = \dfrac{BC}{AC}$

$\dfrac{BD}{40} = \dfrac{30}{50}$

$BD = 24 \, \text{ cm}$

$\sin \beta = \dfrac{y}{BD} = \dfrac{AB}{AC}$

$\dfrac{y}{24} = \dfrac{40}{50}$

$y = 19.2 \, \text{ cm}$ *answer*

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