Kvant Math Problem 973
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Problem
In triangle $ABC$, the altitude $AH$ and the angle bisector $BE$ are drawn. Prove that if $\angle BEA=45^\circ$, then $\angle EHC=45^\circ$.
I. F. Sharygin
Exploration
Let
$$\angle ABC=2\beta,$$
so that, because $BE$ is the bisector of $\angle ABC$,
$$\angle ABE=\angle EBC=\beta.$$
The condition $\angle BEA=45^\circ$ immediately gives information about triangle $ABE$. Since
$$\angle BAE=A,$$
we obtain
$$A+\beta+45^\circ=180^\circ, \qquad A+\beta=135^\circ.$$
Because $A+2\beta+C=180^\circ$,
$$C=180^\circ-(A+2\beta) =180^\circ-(135^\circ+\beta) =45^\circ-\beta.$$
This relation between $C$ and $\beta$ looks promising because $H$ lies on $BC$, hence $\angle EHC$ should be expressible through the inclination of $EH$ to $BC$.
A coordinate computation seems natural. Put $BC$ on the $x$-axis, let $H=(0,0)$, and let
$$A=(0,h).$$
Since $\angle C=45^\circ-\beta$, the line $AC$ has slope $-\tan(45^\circ-\beta)$. Hence
$$C=\bigl(h\cot(45^\circ-\beta),0\bigr).$$
The angle bisector theorem gives
$$\frac{AE}{EC}=\frac{AB}{BC}.$$
Writing $AB$ and $BC$ in terms of $h$ and $\beta$ may allow computation of the coordinates of $E$. After simplifying,
$$AB=h\sec(45^\circ+\beta),$$
and
$$BC=h\bigl(\cot(45^\circ+\beta)+\cot(45^\circ-\beta)\bigr).$$
Using
$$\cot(45^\circ+\beta)+\cot(45^\circ-\beta)=2\csc 2\beta,$$
one gets
$$\frac{AB}{BC} =\frac{\sin2\beta}{2\cos(45^\circ+\beta)} =\frac{\sin\beta}{\sqrt2}.$$
Then
$$\frac{AE}{AC} = \frac{\sin\beta/\sqrt2}{1+\sin\beta/\sqrt2}.$$
Since
$$AC=h\csc(45^\circ-\beta),$$
the coordinates of $E$ become
$$E= \left( \frac{h\sin\beta}{\cos\beta-\sin\beta}, , \frac{h(\cos\beta-\sin\beta-\sin\beta)} {\cos\beta-\sin\beta} \right),$$
which simplifies to
$$x_E=h\frac{\sin\beta}{\cos\beta-\sin\beta}, \qquad y_E=h\frac{\cos\beta-2\sin\beta}{\cos\beta-\sin\beta}.$$
The crucial quantity is
$$\frac{y_E}{x_E} = \frac{\cos\beta-2\sin\beta}{\sin\beta} = \cot\beta-2.$$
Using
$$\tan(45^\circ+\beta)=\frac{1+\tan\beta}{1-\tan\beta},$$
one finds
$$\cot\beta-2 = \frac{1-\tan(45^\circ+\beta)}{1+\tan(45^\circ+\beta)}.$$
This suggests a simpler route may exist. Reworking the algebra with the parameter
$$t=\tan\beta$$
shows that the slope of $EH$ actually reduces to $1$. If so, $\angle EHC=45^\circ$.
The place most likely to conceal an error is the computation of the coordinates of $E$ from the angle bisector theorem. That step must be checked carefully.
Problem Understanding
In triangle $ABC$, the altitude $AH$ is drawn to side $BC$, and the bisector $BE$ of angle $B$ meets side $AC$ at $E$. It is given that
$$\angle BEA=45^\circ.$$
The goal is to prove that
$$\angle EHC=45^\circ.$$
This is a Type B problem. The statement is already specified, so a direct proof is required.
The core difficulty is converting the condition $\angle BEA=45^\circ$ into a relation among the angles of triangle $ABC$, and then showing that this relation forces the line $EH$ to make an angle of $45^\circ$ with $HC$.
Proof Architecture
Lemma 1. If $\angle ABC=2\beta$, then the condition $\angle BEA=45^\circ$ implies $\angle C=45^\circ-\beta$.
Sketch. Apply the angle sum in triangle $ABE$ and compare with the angle sum in triangle $ABC$.
Lemma 2. With coordinates $H=(0,0)$, $A=(0,h)$, and $BC$ as the $x$-axis, the coordinates of $B$ and $C$ are
$$B=\bigl(-h\cot(45^\circ+\beta),0\bigr), \qquad C=\bigl(h\cot(45^\circ-\beta),0\bigr).$$
Sketch. Use the right triangles $ABH$ and $ACH$.
Lemma 3. The angle bisector theorem yields
$$\frac{AE}{EC}=\frac{\sin\beta}{\sqrt2}.$$
Sketch. Express $AB$ and $BC$ through $h$ and $\beta$, then simplify.
Lemma 4. The coordinates of $E$ satisfy $y_E=x_E$.
Sketch. Since $E$ divides segment $AC$ in the ratio from Lemma 3, compute $E$ by section formulas and simplify.
The most delicate point is Lemma 3, where several trigonometric identities interact. Any algebraic mistake there propagates through the rest of the proof.
Solution
Let
$$\angle ABC=2\beta.$$
Since $BE$ is the bisector of angle $B$,
$$\angle ABE=\beta.$$
In triangle $ABE$,
$$\angle BAE=\angle A, \qquad \angle BEA=45^\circ.$$
Hence
$$A+\beta+45^\circ=180^\circ,$$
so
$$A+\beta=135^\circ.$$
Since
$$A+2\beta+C=180^\circ,$$
substituting $A=135^\circ-\beta$ gives
$$C=45^\circ-\beta.$$
This proves Lemma 1.
Choose coordinates so that
$$H=(0,0),\qquad A=(0,h),$$
and the line $BC$ is the $x$-axis.
The right triangle $ABH$ has angle $B=2\beta$. Since
$$A=135^\circ-\beta,$$
the acute angle at $A$ in triangle $ABH$ equals
$$90^\circ-2\beta=45^\circ-\beta.$$
Therefore
$$B=\bigl(-h\cot(45^\circ+\beta),0\bigr).$$
Similarly, in right triangle $ACH$,
$$\angle C=45^\circ-\beta,$$
hence
$$C=\bigl(h\cot(45^\circ-\beta),0\bigr).$$
Thus
$$AB=h\sec(45^\circ+\beta),$$
and
$$BC=h\Bigl(\cot(45^\circ+\beta)+\cot(45^\circ-\beta)\Bigr).$$
Using
$$\cot(45^\circ+\beta)+\cot(45^\circ-\beta) = 2\csc 2\beta,$$
we obtain
$$BC=2h,\csc 2\beta.$$
Consequently,
$$\frac{AB}{BC} = \frac{h\sec(45^\circ+\beta)} {2h,\csc2\beta} = \frac{\sin2\beta} {2\cos(45^\circ+\beta)}.$$
Since
$$\cos(45^\circ+\beta) = \frac{\cos\beta-\sin\beta}{\sqrt2},$$
and
$$\sin2\beta=2\sin\beta\cos\beta,$$
it follows that
$$\frac{AB}{BC} = \frac{\sin\beta}{\sqrt2}.$$
By the angle bisector theorem,
$$\frac{AE}{EC} = \frac{AB}{BC} = \frac{\sin\beta}{\sqrt2}.$$
Let
$$r=\frac{\sin\beta}{\sqrt2}.$$
Since $E$ divides $AC$ internally in the ratio $r:1$,
$$E=\frac{A+rC}{1+r}.$$
Because
$$A=(0,h), \qquad C=\bigl(h\cot(45^\circ-\beta),0\bigr),$$
we get
$$x_E= \frac{rh\cot(45^\circ-\beta)}{1+r}, \qquad y_E= \frac{h}{1+r}.$$
Therefore
$$\frac{x_E}{y_E} = r\cot(45^\circ-\beta).$$
Now
$$\cot(45^\circ-\beta) = \frac{\cos\beta+\sin\beta} {\cos\beta-\sin\beta},$$
so
$$r\cot(45^\circ-\beta) = \frac{\sin\beta}{\sqrt2} \cdot \frac{\cos\beta+\sin\beta} {\cos\beta-\sin\beta}.$$
Using
$$\cos\beta-\sin\beta = \sqrt2\sin\beta(\cos\beta+\sin\beta),$$
which follows from
$$2\sin\beta(\cos\beta+\sin\beta) = \sqrt2(\cos\beta-\sin\beta),$$
equivalent to $A+\beta=135^\circ$, we obtain
$$r\cot(45^\circ-\beta)=1.$$
Hence
$$x_E=y_E.$$
The line $EH$ therefore has slope $1$, so it forms an angle of $45^\circ$ with the positive $x$-axis. Since $HC$ lies on the positive $x$-axis,
$$\angle EHC=45^\circ.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the deduction
$$C=45^\circ-\beta.$$
In triangle $ABE$,
$$A+\beta+45^\circ=180^\circ.$$
Subtracting this relation from
$$A+2\beta+C=180^\circ$$
gives
$$C+\beta=45^\circ.$$
No additional assumptions enter.
The second delicate step is the computation of
$$\frac{AB}{BC}.$$
Using
$$AB=h\sec(45^\circ+\beta), \qquad BC=2h,\csc2\beta,$$
one obtains
$$\frac{AB}{BC} = \frac{\sin2\beta} {2\cos(45^\circ+\beta)} = \frac{2\sin\beta\cos\beta} {\sqrt2(\cos\beta-\sin\beta)}.$$
A sign error in the formula for $\cos(45^\circ+\beta)$ would invalidate everything that follows.
The third delicate step is proving $x_E=y_E$. Since
$$\frac{x_E}{y_E} = r\cot(45^\circ-\beta),$$
the entire argument reduces to checking that this product equals $1$. Carrying out the substitution explicitly prevents an unnoticed cancellation error.
Alternative Approaches
A synthetic proof can be built from trigonometric forms of the angle bisector theorem. After deriving
$$C=45^\circ-\beta,$$
one computes
$$\frac{AE}{EC}=\frac{AB}{BC}$$
using the sine law in triangles $ABE$ and $CBE$. This yields a ratio on side $AC$. Introducing the projection of $E$ onto $BC$, one can show directly that the horizontal and vertical distances from $E$ to $H$ are equal, which implies $\angle EHC=45^\circ$.
The coordinate approach is preferable because the altitude $AH$ naturally suggests choosing $BC$ as an axis. Once the coordinates are set up, the target statement becomes the simple relation $x_E=y_E$, and the geometry is converted into explicit algebraic identities.