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.

Guest Mar 31, 2022

#1**+2 **

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

CPhill Apr 1, 2022