Similarity
In isosceles $\bigtriangleup ABC (AB = AC), CB $ is extended through $B$ to $P$. A line form $P$ , parallel to altitude $BF$, meets $AC$ at $D$ (where $D$ is between $A$ and $F$). From $P$, a perpendicular is drawn to meet the extension of $AB$ at $E$ so that $B$ is between $E$ and $A$. Express $BF$ in terms of $PD$ and $PE$.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
- Charles Caleb Colton
- Charles Caleb Colton
-
Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Re: Similarity
Obvious yet nice problem.
Reflect $D$ and $F$ wrt line $BC$. We get $Y$ and $X$. Now, $\angle ABC=\angle ACB=\angle BCX=\angle BCY$, so $YC$ is parallel to AB. So $PE$ meets $CY$ at $Y$ and $BX\perp CY$. From $CD=CY$, we get $\angle DCP=\angle YCP=\angle EBP= \theta$.
So $PC=\dfrac{PD}{\sin \theta}$ and $PB=\dfrac{PE}{\sin \theta}$. So $BC=PC-PB$ and $BF=BC\sin \theta$. And so $BF=PD-PE$.
Reflect $D$ and $F$ wrt line $BC$. We get $Y$ and $X$. Now, $\angle ABC=\angle ACB=\angle BCX=\angle BCY$, so $YC$ is parallel to AB. So $PE$ meets $CY$ at $Y$ and $BX\perp CY$. From $CD=CY$, we get $\angle DCP=\angle YCP=\angle EBP= \theta$.
So $PC=\dfrac{PD}{\sin \theta}$ and $PB=\dfrac{PE}{\sin \theta}$. So $BC=PC-PB$ and $BF=BC\sin \theta$. And so $BF=PD-PE$.
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.