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 $C$ .

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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.

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    $\begingroup$ 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: $\endgroup$ – Franklin Pezzuti Dyer Sep 27 '20 at 5:23
  • $\begingroup$ @FranklinPezzutiDyer The proof by Christian Blatter below is very clean and simple. $\endgroup$ – Servaes Sep 27 '20 at 14:12
  • $\begingroup$ Here mathoverflow.net/questions/374762 is a generalization of your generalization $\endgroup$ – Đào Thanh Oai Nov 6 '20 at 4:00

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


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

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




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 $AB$.

  • $\begingroup$ In the second line...should it be $BD$ instead of $ID$ ? $\endgroup$ – Peđa Terzić Sep 27 '20 at 6:06
  • $\begingroup$ Yes indeed. I'll change that. $\endgroup$ – 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 $$H={1\over2}(F+G)={1\over2}(A+B)+{\lambda i\over2}(B-A)\ .$$

  • $\begingroup$ +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 $0$. $\endgroup$ – 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 $C$.

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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 $\overrightarrow {BG} = kc \hat i + k(a-b) \hat j$.


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

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.

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    $\begingroup$ 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 $\endgroup$ – Blue Sep 27 '20 at 7:28

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