Square in a Right Triangle: What is this about?
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Copyright © 1996-2012 Alexander BogomolnyThe applet suggests the following statement:
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In DABC angle C is right; F lies on the hypotenuse AB, and K and L on the legs BC and AC such that CKFL is a square. Let CD be the altitude to AB. Then DK and DL are angle bisectors in triangles BCD and ACD. The converse is also true. If CD is the altitude and DK and DL angle bisectors as before, then CKFL is a square. |
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We make two observations:
Since CKFL is a square, the diagonal CF is the bisector of angle C.
Since CD is the altitude to the hypotenuse, triangles ABC, BCD and ACD are similar.
Further, the sides of the square are parallel to the legs of DABC. It follows that, say,
| BK / BF = BC / AB, |
or
| BC / BK = AB / BF. |
From which
| (1) | CK / BK = AF / BF. |
But, in DABC, CF is the bisector of angle C, so that F divides AB in the ratio of the legs:
| (2) | AF / BF = AC / BC. |
Now since triangles ABC and BCD are similar,
| (3) | AC / BC = CD / BD. |
Using transitivity on (1-3), we obtain the proportion
| CK / BK = CD / BD, |
which tells us that DK is the bisector of angle D. Triangle ACD is treated similarly. The argument is reversible.
(This proof is available as a fill-it-in outline.)
References
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Copyright © 1996-2012 Alexander Bogomolny| 40619556 |
