On the sides $CA$ and $CB$ of an isosceles right-angled triangle $ABC$, points $D$ and $E$ are chosen such that $|CD|=|CE|$. The perpendiculars from $D$ and $C$ on $AE$ intersect the hypotenuse $AB$ in $K$ and $L$ respectively. Prove that $|KL|=|LB|$.
Proposed by Victors Linis, University of Ottawa.
Crux Mathematicorum Vol. 1, No. 4, June, 1975
I want a solution via vectors and I'll explain why in the end of the question, tl;dr.
The question consists of:
- the basic things we can do with vectors,
- how did I come to a regular geometric solution,
- regular geometric solution,
- motivation for vectors approach.
To give more explicit context, I'll explain the basic things we can do with vectors to approach real geometry problems.
- We can add or subtract vectors, e.g.
$\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$. - We can scale a vector by a coefficient (say $k$) so if $A,B,C$ lie on the same line and $k=\frac{AC}{AB}$ then $\overrightarrow{AC}=k\overrightarrow{AB}$.
- In particular, 1. and 2. follows that if $X$ is on $AB$, such that $\frac{AX}{XB}=\frac{t}{1-t}$ then $\overrightarrow{OX}$ $=\overrightarrow{OA}+\overrightarrow{AX}$ $=
\overrightarrow{OA}+t\,\overrightarrow{AB}$ $=
\overrightarrow{OA}+t(\overrightarrow{OB}-\overrightarrow{OA})$$=
t\,\overrightarrow{OB}+(1-t)\,\overrightarrow{OA}$. - If some vectors form a basis, then every vector has unique representation as a linear combination of basis vectors with coefficients called "coordinates" (e.g. $\overrightarrow{i},\,\overrightarrow{j},\,\overrightarrow{k}$ is a classical basis for 3d Cartesian coordinates).
Knowing only 1.-4. some problems like this (not in a way attention drawing) may be solved when a convinient basis is chosen, and even Ceva's_theorem, Menelaus's theorem, Thales' theorem can be proven, almost in an algebraic way. I'd call such "linear vector problems". But we know also - Scalar (dot) product. By definition
$\cos\angle BAC=\frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{
|\overrightarrow{BA}|\cdot|\overrightarrow{BC}|}$, or, alternatively, $(\overrightarrow{BA}\cdot\overrightarrow{BC})=BA\cdot BC\cdot \cos\angle BAC$. This implies such things like $(\overrightarrow{BA}\cdot \overrightarrow{BA})=(\overrightarrow{BA})^2=|\overrightarrow{BA}|^2=BA^2$ and $(\overrightarrow{BA}\cdot\overrightarrow{BC})=0\Leftrightarrow BA\perp BC$ unless $BA$ or $BC$ equals to zero. All distributive laws holds for addition/subtraction related to scalar or/and dot product.
With 1.-5. such things like cosine rule, Heron's formula, Ptolemy's_theorem can be proven and I believe the problem above can be solved too.) We also know (though it's usage mostly limited by 3d Cartesian space) - Cross product
Having these tools, we can approach problems, where all conditions given and things to be proven/found are: parallelity, perpendicularity, fixed angles, intersection, intersection at a ratio (and maybe some others). But apparently we can't deal with circles, addition/subtraction of angles and many other things. But boiling a geometric problem down to algebra can be useful when no other ways seen. Other approaches are complex numbers or Cartesian coordinates, but vectors are unfairly less popular/known. I'd say, many vectors excercises are constructed just to train using vectors, instead of showing how real geometrical problems may be solved in an algebraic way.
Arriving at regular geometrical solution
I made the figure above in geogebra and started moving free point $D$ back and forth and see how the things change and I noted that somewhat asymmethrical that we have $3$ points on $AB$ and only two on $AE$, I wanted inverse-image of $B$ to be present to. To construct it, I mirrored $B$ relative to $AE$ into $B'$.
By moving $D$ I noted that $BB'||CL||DK$ (and indeed, they all are perpendicular to $AE$) and that reminded me of Thales' theorem — if we have say $F=BB'\cap AC$
then it would suffice to show that $DC=CF$ and use Thales' theorem. By "method of gazely staring" I found that $\triangle CFB\sim \triangle HEC$, but it's obvious that $\triangle HEC\sim\triangle CEA$, but $CA=CB$ and thus $CE=CF$, but it's given that $CD=CE$, which completes the proof.
Geometrical solution, refined
We take $F$ on the line $AC$ such that $BF||CL$.
$\angle FBC=\angle ECH$, where $H=CL\cap EA$.
From right-angled $\triangle ECH$: $\angle ECH=90^\circ -\angle CEH$,
but from right-angled $\triangle ECA$: $\angle CAE=90^\circ -\angle CEH$
thus $\angle FBC=\angle ECH=\angle EAC$
hence $\triangle FBC$ and $\triangle EAC$ are congruent by ASA
that follows $CF=CE$,
but it's given that $CD=CE$ thus $CF=CD$
and using Thales' theorem on lines $AB$, $AC$ intersected by $BF \parallel CL \parallel DK$ we obtain $BL=LK$, QED.
But imagine I were at a contest without being able to use geogebra and move the point $D$ and to want to construct $BB'$, then arriving at this solution with such additional constructions is highly doubtful. While vectors approach is pretty straightforward: algebraically express what's given and what's needed, solve algebraical problem, usually a linear equations system. That's why I want vectors solution. Other algebraical solutions, like Cartesian coordinates, complex coordinates or even something like barycentric coordinates are welcome as well.
Thanks for reading this through out.)
Best Answer
Rotate the triangle $ABC$ clockwise $90^\circ$ around the point $C$. Then $A$ goes into $A'\!\in \ C\vee B$, and $E$ into $D$. From the following figure it is evident that $|KL|=|LB|$.