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# i am stuck with this question.

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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

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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

Apr 1, 2022