MathLabs

Problem 2

In an acute-angled triangle ABCABC, the interior bisector of angle AA intersects BCBC at LL and intersects the circumcircle of ABCABC again at NN. From point LL, perpendiculars are drawn to ABAB and ACAC, with feet KK and MM respectively. Prove that the quadrilateral AKNMAKNM and the triangle ABCABC have equal areas.
Step 5 of 5: Substitute and recognize the area of ABC
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In plain words

All the auxiliary length ALAL cancels out, leaving precisely the familiar 12⋅(side)⋅(side)⋅sin⁡(included angle)\tfrac12\cdot(\text{side})\cdot(\text{side})\cdot\sin(\text{included angle}) formula for [ABC][ABC] itself — the two areas were the same expression in disguise.

[AKNM]=12⋅AB⋅ACAL⋅ALsin⁡A=12AB⋅ACsin⁡A=[ABC][AKNM] = \frac{1}{2}\cdot\frac{AB\cdot AC}{AL}\cdot AL\sin A = \frac{1}{2}AB\cdot AC\sin A = [ABC]
Detailed analysis

Using AN=AB⋅AC/ALAN=AB\cdot AC/AL from Step 2 and KM=ALsin⁡AKM=AL\sin A from Step 3 in the formula from Step 4: [AKNM]=12⋅AB⋅ACAL⋅ALsin⁡A=12AB⋅ACsin⁡A[AKNM]=\tfrac12\cdot\frac{AB\cdot AC}{AL}\cdot AL\sin A=\tfrac12 AB\cdot AC\sin A. This last expression is exactly the standard two-sides-included-angle formula for the area of △ABC\triangle ABC, so [AKNM]=[ABC][AKNM]=[ABC], as required.