Prove that if $\angle KCL \leq \frac{1}{2}\angle ACB$ then follows that $KL \leq \frac{1}{2}AB$

Question

In an isosceles triangle $ABC$ where $AC = BC$ on a side $AB$ points $K, L$ are chosen so that $\angle KCL \leq \frac{1}{n}\angle ACB$. Prove that for

a) $n = 2$: $KL \leq \frac{1}{2}AB$

b) $n = 3$: $KL \leq \frac{1}{3}AB$

I tried solving a). First we take a look at extreme case: $K \equiv A$ and $\angle KCL = \frac{1}{2}\angle ACB$, then $KL = \frac{1}{2}AB$. Obviously if we decrease $\angle KCL$ length of $KL$ decreases. However I need to prove that if we move $K$ on $AB$ with $\angle KCL = \frac{1}{2}\angle ACB$ length of $KL$ also decreases.

Answer 1

Partial answer: Feels like this approach should be extendable to part b, but it's just not clicking for me.

Proving a)

Consider the perpendicular bisector from $C$ to $AB$ (since it's isosceles) at point $D$. If $K, L$ are on the same side of $D$, then automatically $KL \leq AD = DB = \frac{1}{2} AB$ and $\angle KCL \leq \angle ACD = \angle DCB = \frac{1}{2} \angle ACB$. So we only need to worry if they're on opposite sides.

For ease of algebra, let $h \equiv CD,\; \alpha \equiv \angle ACD,\; \theta \equiv \angle KCL,\; \phi \equiv \angle KCD$. With basic trigonometry, we can get the following, $$ \frac{1}{2} AB = AD = h \tan \alpha \\ KL = KD + DL = h \tan \phi + h \tan (\theta - \phi) = h \tan \theta (1 - \tan \phi \tan (\theta - \phi)) \qquad [\text{using the }\tan(a+b)\text{ formula}] $$

Since $ABC$ is a triangle and $\tan$ is monotonically increasing from $0$ to $\frac{\pi}{2}$, \begin{align} &\angle ACB < \pi \\ &\Rightarrow \quad 0 \leq \phi, \theta - \phi \leq \theta \leq \alpha < \frac{\pi}{2} \\ &\Rightarrow \quad 0 \leq \tan \phi, \tan(\theta - \phi) \leq \tan \theta \leq \tan \alpha < \infty \\ &\Rightarrow \quad 1 - \tan \phi \tan (\theta - \phi)) \leq 1 \end{align}

Now we take the ratio, \begin{align} \frac{KL}{\frac{1}{2} AB} &= \frac{\tan \theta}{\tan \alpha} (1 - \tan \phi \tan (\theta - \phi)) \\ &\leq \frac{\tan \theta}{\tan \alpha} \\ &\leq \frac{\tan \alpha}{\tan \alpha} = 1 \\ &\therefore \quad KL \leq \frac{1}{2} AB \end{align}

Answer 2

A possible path you can follow.

Let $K_1$, $M$, $L_1$ divide $AB$ into $4$ equal parts. Suppose $$KL \cong \frac12 AB.$$ If $K \equiv A$ clearly $KCL \cong \frac12 ACB$. Now, if we show that when $K$ moves from $A$ to $K_1$ the angle $\angle KCL$ increases, then assertion a) is correct, by contradiction (which is just equivalent to what you stated).

In the Figure above, we want to show that $\angle KCL < \angle K_1CL_1$. Consider the point $K'$ symmetric to $K$ w.r.t. $M$. Note that $L_1$ is the midpoint of $LK'$, and since $CL< CK'$, Internal Bisector Theorem on $LCK'$ shows that $\angle LCL_1 > \angle L_1CK' \cong \angle KCK_1$, hence we get $$\angle K_1CL_1\cong \angle K_1CL + \angle LCL_1> \angle K_1CL + \angle KCK_1 \cong\angle KCL,$$ and the assertion is proved. Thus, by contradiction, a) is true.

You can use the same approach for point b), by taking, e.g., $K_1$ and $L_1$ so that they divide $AB$ into three equal parts, and use again Internal Bisector Theorem to show the same behavior of $\angle KCL$, when $K$ moves from $A$ to $K_1$, with $KL \cong \frac13 AB$.

Answer 3

Let's consider the triangle $ABC$ and the point $K$ on the side $AC$ such that $\angle KCL \leq \frac{1}{2}\angle ACB$. We want to prove that $KL \leq \frac{1}{2}AB$. Let $O$ be the circumcenter of triangle $ABC$. Since $O$ is the center of the circumcircle, $OA = OB = OC$ (radii of the circumcircle). Extend $KL$ to intersect the circumcircle at a point $D$. You can observe that $\angle ADB$ is an inscribed angle subtended by the chord $AB$. By the inscribed angle theorem, $\angle ADB = 2 \angle ACB$. Since $\angle KCL \leq \frac{1}{2}\angle ACB$, we have $\angle KCL \leq \angle ADB$. In a circle, if an inscribed angle is larger than another inscribed angle, then the arc opposite the larger angle is longer. Therefore, $KL$ is smaller than the arc length of $AB$ in the circumcircle. $$ KL \leq \text{arc length of } AB = \frac{1}{2}AB$$

Answer 4

By contrapositive, the OP's statements are equivalent to

$(1)$ $KL \gt \frac{1}{2}AB \implies \angle KCL \gt \frac{1}{2} \angle ACB$,

$(2)$ $KL \gt \frac{1}{3}AB \implies \angle KCL \gt \frac{1}{3} \angle ACB$.

It turns out that statement $(1)$ $(n=2)$ is true and statement $(2)$ $(n=3)$ is false.

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Let us first show that the case $n=3$ is false by a counter example:

In the figure, $\Delta ABC$ is an equilateral triangle with $AB=BC=CA=3$. Let $K$ coincide with $A$ and $KL=1.01$ such that $KL \gt \frac{1}{3}AB$.

By cosine rule, \begin{align} CL^2 &= 3^2+1.01^2-2(3)(1.01)\cos 60^{\text o} \\ CL &=2.643880 \\ \end{align} By sine rule, \begin{align} \frac{\sin \angle KCL}{1.01} &=\frac{\sin 60^{\text o}}{CL} \\ \sin \angle KCL &=\frac{1.01 \times \sin 60^{\text o}}{2.643880} \\ \angle KCL &=19.31941^{\text o} \end{align}

$\because \frac{1}{3}\angle ACB=20^{\text o}$

$\therefore \angle KCL \lt \frac{1}{3}\angle ACB $

$\therefore KL \gt \frac{1}{3}AB$ does not imply $\angle KCL \gt \frac{1}{3}\angle ACB$.

The claim for $n=3$ is therefore false.

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Next we are going to prove that the case for $n=2$ is true.

i.e. $KL \gt \frac{1}{2}AB \implies \angle KCL \gt \frac{1}{2} \angle ACB$.

We first state an obvious lemma:

In the figure, $\Delta XYZ$ is a triangle in which $XY, XZ$ have fixed lengths. If $\theta$ varies between $0^{\text o}$ and $180^{\text o}$, then the greater the value of $\theta$, the greater the length of $YZ$.

The converse is also true.

Now for the proof:

In the figure, $D$ is constucted so that $BD=AK$, $CD=CK$.

Note that $$\Delta CAK \cong \Delta CBD$$ Hence $$\angle ACK=\angle BCD$$

Note that $KL \gt \frac{1}{2}AB \implies KL \gt AK+LB$

$\therefore KL \gt LB + BD$

This in turns implies that $KL \gt LD$.

Since $CL=CL$ and $CK=CD$, $KL \gt LD$,

by our lemma, $\angle KCL \gt \angle LCD$.

Noting that $\angle KCL + \angle LCD = \angle ACB$.

This implies that $\angle KCL \gt \frac{1}{2} \angle ACB$.

This complete the proof.