Trapezoid ABCD has parallel bases AB and CD Let AD=6 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
here's what I've done.
+128475 I'm assuming that, as in your drawing, AD = 5 (instead of 6)
We can let AM and BN be the equal height (s) of the tapezoid
So triangles AMD and BNC are right
And triangles PXD and PYC are right
And angle AMD = angle PXD ( since both = 90° )
And angle BNC = angle PYC ( since both = 90° )
And angle ADM = angle XDP
And angle BCN = angle YCP
So, by AA congruency....triangle AMD is similar to triangle PXD
And triangle BNC is similar to triangle PYC
So
AM / AD = PX / PD (1)
And
BN / BC = PY / PC (2)
Let PD = (5/12)DC and PC = (7/12)DC
So, by (1)
h / 5 = PX / [(5/12)DC ] ⇒ h/ 5 = (12/5) PX / DC ⇒ h = 12PX / DC
And by (2)
h/ 7 = PY / [ (7/12)DC ] ⇒ h / 7 = ( 12/7) PY / DC ⇒ h = 12PY/DC
Which implies that
PX / DC = PY /DC
Which implies that
PX = PY
