Triangle Solving – Problem 7: find

In $\triangle ABC$, $AP$ bisects $\angle BAC$ and meets $BC$ at $P$; $BQ$ bisects $\angle ABC$ and meets $CA$ at $Q$. Given $\angle BAC=60^{\circ}$ and $AB+BP=AQ+QB$, find $\angle ABC$.

Step 1. Let $\angle ABC=\theta$, and let the sides opposite $A,B,C$ be $a,b,c$.

Step 2. Using area decomposition $S_{\triangle ABC}=S_{\triangle ABQ}+S_{\triangle CBQ}$, we get $$\frac12 ac\sin\theta=\frac12 c\cdot BQ\sin\frac{\theta}{2}+\frac12 a\cdot BQ\sin\frac{\theta}{2}.$$

Step 3. Solving gives $BQ=\frac{2ac}{a+c}\cos\frac{\theta}{2}$.

Step 4. In $\triangle ABQ$, the Law of Sines gives $\frac{AB}{\sin\angle AQB}=\frac{AQ}{\sin\angle ABQ}$. Similarly, in $\triangle CBQ$, $\frac{BC}{\sin\angle CQB}=\frac{CQ}{\sin\angle CBQ}$.

Step 5. Hence $\frac{AB}{AQ}=\frac{\sin\angle AQB}{\sin\angle ABQ}$ and $\frac{BC}{CQ}=\frac{\sin\angle CQB}{\sin\angle CBQ}$.

Step 6. Since $BQ$ bisects $\angle ABC$, $\angle ABQ=\angle CBQ$. Also, $A,Q,C$ are collinear, so $\angle AQB+\angle CQB=180^{\circ}$, and thus $\sin\angle AQB=\sin\angle CQB$.

Step 7. Therefore $\frac{AB}{AQ}=\frac{BC}{CQ}$, which implies $AQ=\frac{bc}{a+c}$.

Step 8. Similarly, by the angle-bisector theorem, $BP=\frac{ac}{b+c}$.

Step 9. The condition $AB+BP=AQ+QB$ becomes $$c+\frac{ac}{b+c}=\frac{bc}{a+c}+\frac{2ac}{a+c}\cos\frac{\theta}{2}.$$

Step 10. Together with the Law of Cosines $a^{2}+c^{2}-b^{2}=2ac\cos\theta$, this simplifies to $\theta=80^{\circ}$.

Step 11. Hence $\angle ABC=80^{\circ}$.

$$80^{\circ}$$