# 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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Solution 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.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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)

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

1. V. V. Prasolov, Problems in Planimetry, v 1, Nauka, Moscow, 1986, in Russian 