## Outline Mathematics

Geometry

# Square in a Right Triangle

Consider the following problem:

In ΔABC 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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Copyright © 1996-2018 Alexander BogomolnyIn ΔABC 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.

We'll have a sequence of proportions. First of all, since the sides of square CKFL are parallel to the legs,legs,hypotenuse,medians of triangle ABC, we find that, say,

BK / BF = BC / AB,AC,AB,CD,CK,BF,

which is better rewritten as

BC / BK,AC,BK,CK,BF = AB / BF,AC,BK,CK,BF.

And this is equivalent to

(1)

CK,AC,BK,CK,BF / BK = AF / BF.

But in triangle ABC, CF is the bisector,bisector,median,altitude,perpendicular bisector,symmedian of angle C, which implies that F divides AB,AB,BK,CK,BF in the ratio of the legs:

(2)

AF / BF,AB,BK,CK,BF = AC / BC.

Now, since triangles ABC and BCD are similar,

(3)

AC / BC = CD,AB,AC,BD,CD,AD / BD,AB,AC,BD,CD,AD.

Apply transitivity to (1-3)

CK,AB,BK,CK,BF / BK = CD / BD,AB,AC,BD,CD,AD.

In other words, in triangle BCD, K divides BC in the ratio of the sides CD, BD. This exactly means that DK is the bisector of angle D. Triangle ACD is treated similarly.

### References

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