This is quite easy to do using trigonometry, which also establishes that angles BAP and PAC are respectively 24° and 12°. Was your trigonometry question related to this problem?
I haven't found an elegant geometrical proof yet. Where does this problem come from?
This is quite easy to do using trigonometry, which also establishes that angles BAP and PAC are respectively 24° and 12°. Was your trigonometry question related to this problem?
I haven't found an elegant geometrical proof yet. Where does this problem come from?
Construct the regular pentagon $AEBCD$. Angle $BAC = 36^\circ$.
Construct the perpendicular $CH$ from $C$ to $AE$.
Construct the equilateral triangle $AEQ$. By symmetry, $Q$ lies on $CH$.
Draw the line segment $BQ$.
The angles shown below are easily calculated.
Hence $Q$ coincides with the point $P$ given in the problem and so $AP$ = $BC$.
Construct the regular pentagon $AEBCD$. Angle $BAC = 60^\circ$.
Construct the perpendicular $CH$ from $C$ to $AE$.
Construct the equilateral triangle $AEQ$. By symmetry, $Q$ lies on $CH$.
Draw the line segment $BQ$.
The angles shown below are easily calculated.
Hence $Q$ coincides with the point $P$ given in the problem and so $AP$ = $BC$. View attachment 11005