Ipoteză: ΔJKL, ∡JLK ≡ ∡JKL, KM⊥JL, M∈JL, LN⊥JK, N∈JK
Concluzie: A(ΔLMK) = A(ΔKNL)
Demonstrație: KM⊥JL ⇒ ∡KML = 90°, LN⊥JK ⇒ ∡LNK = 90°
∡JLK ≡ ∡JKL, M∈JL, N∈JK ⇒ ∡MLK ≡ ∡NKL
[tex]\left.\begin{matrix} \measuredangle MLK \equiv \measuredangle NKL \\ KL \equiv KL \end{matrix}\right\} \xrightarrow[I.U.]{criteriul} \Delta KML \equiv \Delta LNK\\[/tex]
[tex]\Rightarrow \boldsymbol{KM \equiv LN}[/tex]
[tex]\Rightarrow \boldsymbol{ML \equiv NK}[/tex]
[tex]\left.\begin{matrix} \mathcal{A}_{\Delta LMK} = \dfrac{KM \cdot ML}{2} \\ \\ \mathcal{A}_{\Delta KNL} = \dfrac{LN \cdot NK}{2} \\ \\ KM \equiv LN \\ ML \equiv NK \end{matrix}\right\} \Rightarrow \mathcal{A}_{\Delta LMK} \equiv \mathcal{A}_{\Delta KNL}[/tex]
q.e.d.
