Let $ ABC $ be a triangle and the point $ D $ is on the segment $ BC $ such that $ AD $ is the interior bisector of $ \angle A $. We stretch $ AD $ such that it meets the circumcircle of $ \Delta ABC $ at $ M $. We draw a line from $ D $ such that it meets the lines $ MB,MC $ at $ P,Q $, respectively ($ M $ is not between $ B,P $ and also is not between $ C,Q $). Prove that $ \angle PAQ\geq\angle BAC $.
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Tags: inequalities, geometry, circumcircle, trigonometry, angle bisector, geometry proposed, Geometric Inequalities
jgnr
21.09.2010 12:49
Suppose WLOG that $P$ lies on the extension of $MB$ beyond $B$ and $Q$ lies on segment $CM$. Assume the contrary, $\angle PAQ<\angle BAC$, or $\angle PAQ+\angle PMQ<180^{\circ}$. So point $A$ lies outside the circle $(PMQ)$. The power of point theorem gives $PD\cdot DQ<AD\cdot DM$ and $AD\cdot DM=BD\cdot DC$. Thus $PD\cdot DQ<BD\cdot DC$, $\frac{PD}{BD}<\frac{CD}{QD}$. Let $\angle BDP=\angle QDC=x$. Therefore $\frac{PD}{BD}=\frac{\sin\angle PBD}{\sin\angle BPD}=\frac{\sin\frac{A}2}{\sin(\frac{A}2-x)}$ and $\frac{CD}{QD}=\frac{\sin CQD}{\sin QCD}=\frac{\sin(\frac{A}2+x)}{\sin\frac{A}2}$. We get $\frac{\sin\frac{A}2}{\sin(\frac{A}2-x)}<\frac{\sin(\frac{A}2+x)}{\sin\frac{A}2}$, $\cos x+\cot(\frac{A}2-x)\sin x<\cos x+\cot\frac{A}2\sin x$, $\cot(\frac{A}2-x)<\cot\frac{A}2$, $\frac{A}2-x>\frac{A}2$, a contradiction. Hence $\angle PAQ\ge\angle BAC$.
mahanmath
21.09.2010 13:54
It has an easier solution , add two circles into your figure ...
siavosh
15.04.2012 17:15
$\angle MBC= \angle MAC=\angle BAM $ it means that $MB$ is tangent to $\bigodot ABD \Rightarrow \angle APD \leq \angle ABC $ (1) similary we can show $ MC $ is tangent to $ \bigodot ADC \Rightarrow \angle AQD \leq \angle ACD$ (2) (1),(2)give us that $\angle PAQ \geq \angle BAC $ solution
MBGO
15.04.2012 21:22
Use Menelaus and Sines Theorem + Angle busector theorem...
MBGO
15.04.2012 21:30
Dear all ================================================================================ here are some tips... ================================================================================
Suppoese that the intersection Point $Q$ is inside the circumcircle of $ABC$ and the other intersection point is outside the circumcircle of $ABC$
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Use menelaus about triangle$MBC$ and the line $PQ$ and then by angle bisector theorem get that : $\frac{AB*MP}{PB}=\frac{AC*QM}{CQ}$
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Suppose $\measuredangle PAB=\alpha$ ,$\measuredangle QAC=\beta$ and $\frac{\measuredangle A}{2}=\gamma$ and by sines Theorem get that :$\frac{sin(\gamma -\beta )}{sin\beta }=\frac{\gamma +\alpha}{sin\alpha }$
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by some angle chasing(about 2 line) get that $sin \alpha \geq sin \beta$ and $\beta \leq 90$ degrees and then finish the problem by using $\alpha \geq \beta$
================================================================================ And we have Done.
Mahdi_Mashayekhi
09.01.2022 15:15
∠BAD = ∠MAC = ∠MBC ---> ABD is tangent to BM at B so P is out of circle ABD. ∠DAC = ∠MAB = ∠MCB ---> ADC is tangent to CM at C so Q is out of circle ACD. ∠DPA < ∠DBA and ∠DQA < ∠DCA so ∠BAC < ∠PAQ. Equality holds only if P is B and Q is C.