Perpendicular on Angle Bisector — Special Case
November 3, 2024 2025-11-03 22:15Perpendicular on Angle Bisector — Special Case
Perpendicular on Angle Bisector — Special Case
In \( \triangle ABC \), let \( AD \) be the internal bisector of \( \angle A \). From \(B\), drop a perpendicular \( BP \perp AD \). Let \( M \) be the midpoint of \( BC \). Then the handy result is \[ \boxed{\,PM=\tfrac{|AC-AB|}{2}\,}. \] (Take absolute value if needed.)
Why the shortcut works
In the configuration, symmetry about the angle bisector makes \( \triangle ABE \) isosceles with \( AB=AE \), hence \(P\) is the midpoint of \(BE\).
With \(M\) the midpoint of \(BC\), the Midpoint Theorem in \( \triangle BEC \) gives \( PM \parallel EC \) and
\( PM = \dfrac{EC}{2} = \dfrac{AC-AB}{2} \).
CAT-style question
Question. In triangle \(ABC\), let \(AB=7~\text{cm}\), \(BC=8~\text{cm}\), and \(AC=13~\text{cm}\). Let \(M\) be the midpoint of \(BC\), and let \(D\in BC\) be such that \(AD\) is the internal angle bisector of \( \angle BAC \). From \(B\), drop a perpendicular \(BP\perp AD\) meeting \(AD\) at \(P\). Find \(PM\) (in cm).
Click for solution
Using the special case, \[ PM=\frac{AC-AB}{2}=\frac{13-7}{2}=\boxed{3}. \]
Quick review
- Drop \(BP\perp\) to the **internal** angle bisector \(AD\).
- Let \(M\) be midpoint of \(BC\).
- Direct read-off: \( PM=\dfrac{|AC-AB|}{2} \).
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