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Prove that in a right triangle $ABC$, the bisector of the right angle bisects the angle formed between the median and the altitude drawn from the same vertex.


In other words, I'm asked to prove that $\angle MBD = \angle DBH$; according to the picture below.

enter image description here

This is the same as proving that $\displaystyle \frac{BM}{BH} = \frac{MD}{DH}$

Others facts are also known:

  • $AM = MB = MC$

  • $\angle ABD = \angle CBD = 45º$

I'm not sure how to use that information. So what I tried to do is to construct a triangle so that I would be able to prove the equality by similarity, but I'm stuck:

enter image description here

$1)$ $DF \perp AB$, by construction.

$2)$ Quadrilateral $BFDH$ is circumscribed by a circle, since $m\angle BFD + m\angle BHD = 180º$.

$3)$ $\angle HFD = \angle DBH$, inscribed angles subtended by the same arc $\stackrel{\frown}{DH}$ (It follows from $2)$).

$4)$ $\angle FDB = \angle FHB$, inscribed angles subtended by the same arc $\stackrel{\frown}{FB}$ (It follows from $2)$).

$5)$ $\triangle FDO \sim \triangle BHO$, by AA similarity.

I don't know how to go further. It would be nice if there is some way to prove that $\angle BPD = \angle BOH$.

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  • $\begingroup$ And if one angle is $\dfrac\pi8$ , then the three lines quadrisect the right angle into four equal pieces. $\endgroup$ – Lucian Oct 15 '15 at 23:58
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Here is the way:

$\angle ABM=\angle CAB$ (isocele triangle), and $\angle HBC=\angle CAB$ as well ($HBC$ $HAB$ similar triangles).

$\angle ABD=\angle DBC$ by definition, thus you have

$\angle DBM=\angle ABD-\angle ABM=\angle DBC-\angle HBC=\angle HBD$, which is what you wanted to prove.

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  • $\begingroup$ Pretty neat. It took me a bit to see how it goes. Thanks. $\endgroup$ – Jazz Jun 11 '15 at 17:06

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