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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.

 

 

 Mar 31, 2022
 #1
avatar+124594 
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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

 

 

cool cool cool

 Apr 1, 2022

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