# A generalization of Bottema's theorem

Can you provide a proof for the following claim:

In any triangle $$ABC$$ construct isosceles right triangles on sides $$AC$$ and $$BC$$, with right angles at the points $$A$$ and $$B$$. Let points $$F$$ and $$G$$ divide catheti $$AE$$ and $$BD$$ respectively in the same arbitrary ratio . The midpoint $$H$$ of the line segment that connects points $$F$$ and $$G$$ is independent of the location of $$C$$ .

GeoGebra applet that demonstrates this claim can be found here. I tried to mimic a proof of Bottema's theorem given on this page but without success.

• It hurts me when I see a fact like this that is so clean and simple but seems to require lots of nasty trig to prove. D: – Franklin Pezzuti Dyer Sep 27 '20 at 5:23
• @FranklinPezzutiDyer The proof by Christian Blatter below is very clean and simple. – Servaes Sep 27 '20 at 14:12
• Here mathoverflow.net/questions/374762 is a generalization of your generalization – Đào Thanh Oai Nov 6 '20 at 4:00

Basically here we mimic the proof from cut-the-knot, but replacing congruence with similarity:

Let

$$\frac {AH}{AG} = \frac {BI}{BD} = a$$

Let's drop perpendiculars $$HL$$, $$CX$$, $$JK$$, and $$IM$$ onto $$AB$$. (I forgot to label $$X$$)

$$JK$$ is the midline of trapezoid $$HLMI$$ so that

$$JK = \frac {HL + IM}2$$

Further, since $$\angle HAC$$ is right, $$\angle HAL$$ and $$\angle CAX$$ are complementary which makes right triangles $$\triangle HAL$$ and $$\triangle ACX$$ similar, implying

$$HL=aAX$$

Similarly,

$$IM=aBX$$

Taking all three identities into account shows that

$$JK = \frac {HL + IM}2 = \frac a2 (AX+BX) = \frac a2 AB = aAK$$

independent of $$C$$. No trigs but tricks.

EDIT: I see how I can prove that $$AK=KB$$.

By the previous similar triangles ($$\triangle HAL \sim \triangle ACX$$ and $$\triangle IBM \sim \triangle BCX$$), we have:

$$AL = k CX = BM$$

By midline theorem, $$LK = KM$$.

Therefore $$AK = LK-AL=KM-BM=KB$$.

This shows that (finally!) $$J$$ is fixed, since it is at a fixed distance "above" the midpoint of $$AB$$.

• In the second line...should it be $BD$ instead of $ID$ ? – Peđa Terzić Sep 27 '20 at 6:06
• Yes indeed. I'll change that. – player3236 Sep 27 '20 at 6:06

Consider $$A$$, $$B$$, $$C$$ as complex numbers, and choose a $$\lambda\in{\mathbb R}$$. Then $$F=A+\lambda(E-A)=A+\lambda\,i(C-A),\qquad G=B+\lambda(D-B)=B+\lambda(-i) (C-B)\ .$$ It follows that $$H={1\over2}(F+G)={1\over2}(A+B)+{\lambda i\over2}(B-A)\ .$$

• +1 This is the way to go. Also note that this shows the same proof works if the two angles at $A$ and $B$ are not required to be right angles, but only required to sum to $0$. – Servaes Sep 27 '20 at 14:10

A little angle-chasing shows that the target point (here, $$K$$) is the midpoint of a side of a particular, symmetrically-situated parallelogram, which in turn shows that, for a given ratio $$\lambda$$, the point's position relative to side $$\overline{AB}$$ is independent of the position of $$C$$.

Note: $$c$$ is half of $$|AB|$$ in the figure.

FYI: If the right angles are formed "the other way" at $$A$$ and $$B$$, then the corresponding midpoint is the reflection of $$K$$ across $$\overline{AB}$$. Proof is left as an exercise to the reader.

Here is a proof via vectors. This avoids the issue of the location of $$J$$ in my earlier proof.

Use the original diagram and let $$O$$ be the midpoint of $$AB$$.

Let $$\overrightarrow {OB} = a \hat i$$. Then $$\overrightarrow {OA} = -a \hat i$$.

Let $$\overrightarrow {OC} = b \hat i + c \hat j$$.

Hence $$\overrightarrow {AC} = (a+b) \hat i + c \hat j$$ and $$\overrightarrow {BC} = (-a+b) \hat i + c \hat j$$.

We can easily show that $$\overrightarrow {AE} = -c \hat i+(a+b) \hat j$$ and $$\overrightarrow {BD} = c \hat i + (a-b) \hat j$$.

Letting $$\dfrac {AF}{AE} = \dfrac {BG}{BD} = k$$ we have $$\overrightarrow {AF} = -kc \hat i+k(a+b) \hat j$$ and $$\overrightarrow {BG} = kc \hat i + k(a-b) \hat j$$.

Finally:

\begin{align}\overrightarrow{OH}&=\frac12(\overrightarrow{OF}+\overrightarrow{OG})\\&=\frac12(\overrightarrow{OA}+\overrightarrow{AF}+\overrightarrow{OB}+\overrightarrow{BG}) \\&=\frac12(-a\hat i-kc \hat i+k(a+b) \hat j+a\hat i+kc \hat i + k(a-b) \hat j) \\&=\frac k2((a+b)+(a-b))\hat j \\&=ka\hat j \end{align}

This shows that $$OH \perp AB$$ and $$|OH|$$ only depend on $$a$$ and $$k$$, that is, the length of $$AB$$ and the ratio $$k$$, implying the position of $$H$$ is indeed fixed.

• Stated differently, let $v^\perp$ be the vector obtained by swapping the components of $v$ and changing one of the signs. Then we have \begin{align} H &= \tfrac12(D+E)\\[4pt] &=\tfrac12((A+k(C-A)^\perp)+(B - k(C-B)^\perp)) \\[4pt] &=\tfrac12((A+B)+k(C-A-(C-B))^\perp)\\[4pt] &=\tfrac12((A+B)+k(B-A)^\perp) \\[4pt] \end{align} which is clearly independent of $C$. Sneakily, I didn't specify which component's sign is changed by $\perp$. Each choice leads to a valid construction of a corresponding fixed point (either "above" or "below" $\overline{AB}$). This proves the "FYI" in my answer – Blue Sep 27 '20 at 7:28