Let $ABC$ be an isosceles triangle with $AB=AC$. Let $D$ be a point on the segment $BC$ such that $BD=2DC$. Let $P$ be a point on the segment $AD$ such that $\angle BAC=\angle BPD$. Prove that $\angle BAC=2\angle DPC$.
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Tags: geometry, trigonometry, geometry unsolved
BBAI
09.06.2012 21:33
Soln: Extend $AD$ to $X$ such that$ BP=PX $. So $\triangle BPX$ is similar to $\triangle ABC$.$ \Longrightarrow $ $X$ lies on $ \odot ABC$. As $BD=2DC$.,so $BX=2XC.$ Let a perpendicular is drawn from $P $to $ BX $.at $Y$. So $XY=XC$. $ \triangle XYC $ is isosceles .$\Longrightarrow $ $\triangle PYC$ is isosceles. So $\angle CPD $ =$\angle XPY$ =$\frac{\angle BAC}{2}.$ So done.
subham1729
09.06.2012 21:38
It was a problem of Indian Imotc 2012 practise test I did there by trigonmetry.
BBAI
09.06.2012 22:49
Everyone knows you solve geometry by Sine rule and coordinate.
sunken rock
09.06.2012 22:59
See http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=316752 as well! Best regards, sunken rock
Goutham
10.06.2012 07:01
Take $K$ the midpoint of $BP$. Simple sine rule gives $BP=2AP=BK$ and prove that $ABK\cong CAP$ and angle chase gives the result.
Nishhari_5
26.03.2018 15:28
BBAI wrote: Soln: Extend $AD$ to $X$ such that$ BP=PX $. So $\triangle BPX$ is similar to $\triangle ABC$.$ \Longrightarrow $ $X$ lies on $ \odot ABC$. As $BD=2DC$.,so $BX=2XC.$ Let a perpendicular is drawn from $P $to $ BX $.at $Y$. So $XY=XC$. $ \triangle XYC $ is isosceles .$\Longrightarrow $ $\triangle PYC$ is isosceles. So $\angle CPD $ =$\angle XPY$ =$\frac{\angle BAC}{2}.$ So done. If $XYC $ is isosceles then we will get quadrilateral $PYXC$ is cyclic which will finish the problem, but how do we get $\Delta PYC$ is isosceles Forget it, I got it