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Question Number 41348 by behi83417@gmail.com last updated on 06/Aug/18

Commented by MJS last updated on 06/Aug/18

$trapezoid A B C D$ $a = A B b = B C c = C D d = D A$ $e = A C f = B D$ $which triangles do you mean ?$ $△ A B C and △ A B D ?$
Commented by behi83417@gmail.com last updated on 06/Aug/18

$t h e r e i s 4 t r i a n g l e s . y o u c a n$ $a s s u m e a n y 2 t r i a n g l e s w i t h$ $a r e a s m & n . 2 o p p o s i t e o r 2 n e i g h b o r$ $t r i a n g l e s . q u e s t i o n h a v e m o r e$ $t h a n 1 a n s e w e r .$
Commented by behi83417@gmail.com last updated on 06/Aug/18

Commented by behi83417@gmail.com last updated on 06/Aug/18

Commented by behi83417@gmail.com last updated on 06/Aug/18

$△ o A B \sim △ o D C \Rightarrow \frac{o A}{o C} = \frac{o B}{o D} = \frac{A B}{D C} = k$ $\frac{S_{o A B}}{S_{o D C}} = k^{2} \Rightarrow \frac{m}{n} = k^{2} \Rightarrow k = \sqrt{\frac{m}{n}}$ $S_{A B C D} = \frac{1}{2} (A B + C D) . A P = \frac{1}{2} (A B + C D) (o H + {o H}^{'}) =$ $= \frac{1}{2} (A B + C D) (\frac{2 m}{A B} + \frac{2 n}{C D}) =$ $= \frac{1}{2} (A B . \frac{2 m}{A B} + A B . \frac{2 n}{C D} + C D . \frac{2 m}{A B} + C D . \frac{2 n}{C D}) =$ $= \frac{1}{2} (2 m + 2 n . \sqrt{\frac{m}{n}} + 2 m . \sqrt{\frac{n}{m}} + 2 n) =$ $= (m + 2 \sqrt{m n} + n) = {(\sqrt{m} + \sqrt{n})}^{2} .$
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