In triangle $ABC$, $BC > AC$. Point $D$ is on side $BC$ such that $BD = BC - AC$. Prove that line $AD$ is perpendicular to the angle bisector of $\angle ACB$.

1. Problem Description

In triangle $ABC$, $BC > AC$. Point $D$ is on side $BC$ such that $BD = BC - AC$. Prove that line $AD$ is perpendicular to the angle bisector of $\angle ACB$.

2. Solution Steps

Let $CE$ be the angle bisector of $\angle ACB$. Let the intersection of $AD$ and $CE$ be point $F$. We want to prove that $AD \perp CE$, which means $\angle AFC = 90^\circ$.

Let $AC = b$ and $BC = a$. We are given that $BD = BC - AC = a - b$. Thus, $CD = BC - BD = a - (a - b) = b$. Therefore, $CD = AC = b$. This means that triangle $ACD$ is isosceles with $CD = AC$. Since $CE$ is the angle bisector of $\angle ACB$, it is also the perpendicular bisector of $AD$ in the isosceles triangle if the triangle is isosceles with $CA=CD$.

Since $CD = AC$, $\triangle ACD$ is an isosceles triangle. Let $\angle CAD = \angle CDA = \alpha$.

Since $CE$ is the angle bisector of $\angle ACB$, we have $\angle ACE = \angle BCE = \frac{\gamma}{2}$, where $\angle ACB = \gamma$.

In $\triangle ABC$, $\angle BAC = \alpha$, $\angle ABC = \beta$, and $\angle ACB = \gamma$. We know that $\alpha + \beta + \gamma = 180^\circ$.

In $\triangle ADC$, $\angle CAD = \angle CDA = \alpha$. Then $\angle ACD = 180^\circ - 2\alpha$.

Since $\angle ACB = \gamma$, we have $\angle BCD = \gamma - (180^\circ - 2\alpha)$.

Since $D$ lies on $BC$, we must have $CD=AC$.

Also $\angle ACB = \gamma$ and $\angle ACE = \angle BCE = \frac{\gamma}{2}$.

Let $K$ be the intersection of the angle bisector of $\angle ACB$ and $AD$. We want to prove $\angle AKC = 90^\circ$.

Since $\triangle ACD$ is isosceles with $AC=CD$, and $CK$ is the angle bisector of $\angle ACB$, it does not mean that $CK$ is perpendicular to $AD$.

Since $CE$ is the angle bisector of $\angle ACB$, we have $\angle ACE = \angle BCE$.

Let's pick point $E$ on $BC$ such that $AC = CE = b$. Since $BC = a$ and $BD = a-b$, we have $DE = BE - BD = (BC - EC) - (BC - AC) = AC - EC = 0$, thus $D=E$. Then $CD=AC$ and hence $\triangle ADC$ is an isosceles triangle. The bisector of $\angle ACD$ is perpendicular to $AD$, and it bisects $AD$. Let $CK$ be the bisector of $\angle ACD$, where $K$ lies on $AD$. $\angle ACK = \angle DCK$.

Since $CD = AC$, $\angle CAD = \angle ADC = \alpha$. $\angle ACD = 180^\circ - 2\alpha$.

Thus $\angle ACK = \angle DCK = (180^\circ - 2\alpha)/2 = 90^\circ - \alpha$.

Also, $\angle AKC = 90^\circ$, i.e. $AD \perp CK$. We need to show that $CE$ coincides with $CK$.

Then $AC = CD = b$, and $\angle ACB = \gamma$. Therefore $\angle ACD = \angle ACB = \gamma$.

Since $\angle ACD + \angle DCB = \angle ACB = \gamma$, it follows that $AC=CD$, so $\angle ACB = \angle ACD$.

Therefore, $CK$ is also the angle bisector of $\angle ACB$. Since $\angle ACK = 90^\circ - \alpha$, $CE$ coincides with $CK$ and $AD \perp CE$.

3. Final Answer

The line $AD$ is perpendicular to the angle bisector of $\angle ACB$.