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In $\triangle ABC$ $BE$ and $AD$ are heights.
Create rectangle $AEPQ$ that $AC = AQ$ and create rectangle $DBNM$ that $BC = BN$.
Show that Area of square $LKBA$ is equal to sum of areas of rectangles.
Thanks for help :) enter image description here

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In $\triangle ABC$, $AD$ and $BE$ are heights. Suppose they meet at $O$. Suppose $CO$ produced meets $AB$ and $LK$ at $R$ and $G$ respectively.

So $CG \parallel BK$ . Join $CK$ and $AN$.

In $\triangle CBK$ and $\triangle ABN$, $$BC=BN\\ AB=BK\\ \angle ABN = \angle CBK= 90^{\circ} + \angle B$$

so $\triangle CBK$ is congruent to $\triangle ABN$.


since parallelogram $BRGK$ & $\triangle CBK$ are on same base $BK$ and $BK \parallel CG$. Equivalently, $(ARGL)=(AEPQ)$. So $$(BRGK) + (ARGL)=(BDMN)+ (AEPQ)=(ABKL)$$

Here $(XYZ)$ or $(ABCD)$ means area of the triangle $XYZ$ or quadrilateral $ABCD$.

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Hint. Say $a, b, c$ are the sides of the triangle, $a$ being $AB$, $h_a, h_b, h_c$ are its heights and $S$ is its area. The equation to prove then is $b\sqrt{a^2-h_b^2} + c\sqrt{a^2 - h_c^2} = a^2$. This follows from Pythagoras. Express $h_i$ through $S$ and see how you could transform them into $h_a$.

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