# Intersection of median and altitude of right angled triangle

In a right angled triangle, we are given that the smaller lengths (i.e. the sides other than the hypotenuse) are $30\text{ cm}$ and $30\sqrt{3}\text{ cm}$. The median to the smallest side (i.e. the line joining the midpoint of the smallest side to its opposite point) and the perpendicular from the right angle to the hypotenuse meet at a point. Determine the distance from the foot of the perpendicular to this point.

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Let $\triangle ABC$ be the right angled triangle, the right angle being $\angle ABC$. Let $BC=30\text{ cm}$, $AB=30\sqrt{3}\text{ cm}$. Then note that the smallest side will be $BC$. Let $D$ be the midpoint of $BC$ and $E$ be the foot of perpendicular from $B$ to $AC$. Let $AD$ and $BE$ meet at $F$. Let $\angle BCA= \alpha$. Then we have \begin{align} \tan(\alpha) &= 30\sqrt{3}/30\\ \tan(\alpha) &= \sqrt{3}\\ \alpha &= 60^\circ \end{align} Now $D$ being the midpoint of $BC$, we have $BD=BC/2=15\text{ cm}$. Now let $\angle BDF=\theta$. Then we have \begin{align} \tan(\theta) &= 30\sqrt{3}/15\\ \tan(\theta) &= 2\sqrt{3} \end{align} Consider another right angled triangle with one of its angles $\theta$, so that the opposite side of $\theta$ will be of length $2\sqrt{3}k$, where $k$ is some real number multiplied by an unit of length. Then the adjacent side of $\theta$ will be $k$. Then according to the Pythagorean Theorem, the length of the hypotenuse will be $\sqrt{13}k$. Then we have $\sin(\theta)=2\sqrt{3}/\sqrt{13}$ and $\cos(\theta)=1/\sqrt{13}$.

Now back to the original problem. Consider the right angled $\triangle BEC$. We get $$\angle EBC + \angle BEC + \angle BCE = 180^\circ \Rightarrow \angle EBC = 30^\circ$$ Now consider $\triangle BFD$. We have $$\angle BFD = 180^\circ - (\angle EBC + \theta) = 180^\circ - (30^\circ + \theta)$$ Apply sine rule to $\triangle BFD$. \begin{align} BD/\sin(180^\circ - (30^\circ + \theta)) &= BF/\sin(\theta)\\ BD/\sin(30^\circ + \theta) &= BF/\sin(\theta)\\ BD/(\sin(30^\circ)\cos(\theta) + \cos(30^\circ)\sin(\theta)) &= BF/\sin(\theta)\\ BD/(1/2\cdot1/\sqrt{13} + \sqrt{3}/2\cdot2\sqrt{3}/\sqrt{13}) &= BF/(2\sqrt{3}/\sqrt{13})\hspace{4cm}\\ BD/((1/2 + 3)/\sqrt{13}) &= BF/(2\sqrt{3}/\sqrt{13})\\ BD/(7/2) &= BF/(2\sqrt{3})\\ BF &= BD\cdot2/7\cdot2\sqrt{3}\\ &= 60\sqrt{3}/7\text{ cm} \tag{since BD=15\text{ cm}} \end{align} Now in the right angled $\triangle BEC$, \begin{align} \sin(\alpha) &= BE/BC\\ BE/BC &= \sqrt{3}/2\\ BE &= 15\sqrt{3}\text{ cm}\tag{since BC = 30\text{ cm}} \end{align} Now we wish to find $EF$, so $EF = BE-BF = (15-60/7)\sqrt{3}\text{ cm} = 45\sqrt{3}/7\text{ cm}$.

This should be our final answer. But they say that my answer is wrong. Can anybody please point out the mistake in my arguments?

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Just to clarify, in the sentence "determine the distance from the foot of the perpendicular to this point", by "the foot of the perpendicular" I mean the foot of the perpendicular from the right angle to the hypotenuse. –  abcd Apr 4 '13 at 9:58
Hey ! Could you copy your previous $3$ comments,i.e., the work you have done in the main post itself and delete these two comments. That would be nice to view. –  lsp Apr 4 '13 at 10:27
@lsp, okay..... –  abcd Apr 4 '13 at 10:30
Check the math on $(15-60/7)\sqrt{3}\text{ cm} = 15\sqrt{3}/7\text{ cm}$. This appears to be the same as this question. Check the answers there. –  robjohn Apr 4 '13 at 11:25
@robjohn Yes made a silly mistake there. Now they say my answer is right. Thanks very much. –  abcd Apr 4 '13 at 16:51