Let
![ABCD](/media/m/9/c/e/9ce25711ba18d9663b73c3580de4bf5a.png)
be a convex quadrilateral such that the sides
![AB, AD, BC](/media/m/5/1/9/5198095d11cf95bb48da2eeffab83295.png)
satisfy
![AB = AD + BC.](/media/m/4/2/d/42dac12b97235f7ab1c7cb1fa07eb315.png)
There exists a point
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
inside the quadrilateral at a distance
![h](/media/m/e/4/3/e438ac862510e579cf5cbdbe5904d4ba.png)
from the line
![CD](/media/m/8/9/5/895081147290365ccae028796608097d.png)
such that
![AP = h + AD](/media/m/3/1/0/3106dabb16d6ef617eebc5eed93850fd.png)
and
![BP = h + BC.](/media/m/4/5/8/458e6fc2dd9daf252c1453b92860412a.png)
Show that:
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Let $ABCD$ be a convex quadrilateral such that the sides $AB, AD, BC$ satisfy $AB = AD + BC.$ There exists a point $P$ inside the quadrilateral at a distance $h$ from the line $CD$ such that $AP = h + AD$ and $BP = h + BC.$ Show that:
$$\frac {1}{\sqrt {h}} \geq \frac {1}{\sqrt {AD}} + \frac {1}{\sqrt {BC}}$$