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Trapezoid ABCD has parallel 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.

 Nov 27, 2020
 #1
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This problem sounds interesting:

 

If we draw PA and PB to make ∆PDA and ∆PCB, we have that the altitudes must be equal, so the areas of PAD and PBC are in the ratio 5:7.

 

If PX is the altitude to AD, we can treat AD as the base instead of PD, thus PX is the altitude and same for BC and PC. Thus, we have that PX*5/2 : PY*7/2 = 5:7, so 5PX : 7PY = 5:7. Therefore, dividing both sides by 5 and then multiplying by 5/7, we get PX : PY = 1:1, so PX = PY.

 Nov 27, 2020
 #2
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Thank you so much for your help! Your solution is so clear and simple! I guess I should have thought of drawing PA & PB! Thanks! smiley 

 Nov 27, 2020

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