Let ABCD be a trapezoid with bases AB and CD. Let AD=5 and BC=7, and let P be a point on side CD such that CP/PD=7/5. Let X,Y be the feet of the altitudes from P to AD,BC respectively. Show that PX=PY.
I don't quite understand how to diagram 'Let X,Y be the feet of the altitudes from P to AD,BC respectively.' Can some one help?
+26367 Let ABCD be a trapezoid with bases AB and CD. Let AD=5 and BC=7, and let P be a point on side CD such that CP/PD=7/5.
Let X,Y be the feet of the altitudes from P to AD,BC respectively.
Show that PX=PY.
\(\text{Let area of $\triangle ADP = A_1$} \\ \text{Let area of $\triangle BCP = A_2$} \)
\(\begin{array}{|rcll|} \hline 2A_1 &=& \dfrac{5}{12}DC*h \\\\ 2A_2 &=& \dfrac{7}{12}DC*h \\\\ \hline \dfrac{2A_1}{2A_2} &=& \dfrac{\dfrac{5}{12}DC*h}{\dfrac{7}{12}DC*h} \\\\ \dfrac{A_1}{A_2} &=& \dfrac{5}{12} * \dfrac{12}{7} \\\\ \mathbf{ \dfrac{A_1}{A_2} } &=& \mathbf{ \dfrac{5}{7} } \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline 2A_1 &=& AD*PX \\\\ 2A_2 &=& BC*PY \\\\ \hline \dfrac{2A_1}{2A_2} &=& \dfrac{ AD*PX } {BC*PY} \quad | \quad AD=5,\ DC=7 \\\\ \dfrac{A_1}{A_2} &=& \dfrac{ 5*PX } {7*PY} \quad | \quad \mathbf{ \dfrac{A_1}{A_2} = \dfrac{5}{7} } \\\\ \dfrac{5}{7} &=& \dfrac{ 5*PX } {7*PY} \\\\ \dfrac{5}{7}*\dfrac{7}{5} &=& \dfrac{PX}{PY} \\\\ 1 &=& \dfrac{PX}{PY} \quad | \quad * PY \\\\ \mathbf{PY} &=& \mathbf{PX} \\ \hline \end{array}\)
