E.90. În triunghiul ABCABCABC cu m(∡B\measuredangle B∡B) >\gt> m(∡C\measuredangle C∡C) avem AD⊥BC,D∈BCAD \bot BC, D\in BCAD⊥BC,D∈BC și EEE mijlocul lui [BC].[BC].[BC]. Dacă AB=2⋅DE,AB = 2 \cdot DE,AB=2⋅DE, arătați că m(∡B)=2⋅m(∡C)m(\measuredangle B) = 2 \cdot m(\measuredangle C)m(∡B)=2⋅m(∡C).
Indicația 1: Construim EFEFEF linie mijl. în △ABC\triangle ABC△ABC
Indicația 2: ∡F1≡∡E1≡∡C\measuredangle F_1 \equiv \measuredangle E_1 \equiv \measuredangle C∡F1≡∡E1≡∡C
Indicația 3: ∡D1=2⋅∡E1\measuredangle D_1 = 2 \cdot \measuredangle E1∡D1=2⋅∡E1
Construim EFEFEF linie mijl. în △ABC⇒EF∥AC⇒∡E1≡∡C (1)\triangle ABC \textcolor{red}{\Rightarrow} EF \parallel AC \textcolor{red}{\Rightarrow} \boxed{\measuredangle E_1 \equiv \measuredangle C} \nobreakspace \textcolor{red}{(1)}△ABC⇒EF∥AC⇒∡E1≡∡C (1) În △ADB,∡D=90°,DF\triangle ADB, \measuredangle D = 90\degree, DF△ADB,∡D=90°,DF mediană ⇒DF=AB2.\textcolor{red}{\Rightarrow} DF = \dfrac{AB}{2}.⇒DF=2AB. Dar și DE=ip.AB2⇒DF≡DE⇒∡E1≡∡F1.DE \xlongequal{ip.} \dfrac{AB}{2} \textcolor{red}{\Rightarrow} DF \equiv DE \textcolor{red}{\Rightarrow} \measuredangle E_1 \equiv \measuredangle F_1.DEip.2AB⇒DF≡DE⇒∡E1≡∡F1.
D1D_1D1 unghi exterior ⇒∡D1=∡E1+∡F1⇒∡D1=2⋅∡E1 (2)\textcolor{red}{\Rightarrow} \measuredangle D_1 = \measuredangle E_1 + \measuredangle F_1 \textcolor{red}{\Rightarrow} \boxed{\measuredangle D_1 = 2 \cdot \measuredangle E1} \nobreakspace \textcolor{red}{(2)}⇒∡D1=∡E1+∡F1⇒∡D1=2⋅∡E1 (2) Din FB≡FD⇒∡B≡∡D1 (3)FB \equiv FD \textcolor{red}{\Rightarrow} \boxed{\measuredangle B \equiv \measuredangle D_1} \nobreakspace \textcolor{red}{(3)}FB≡FD⇒∡B≡∡D1 (3)
Din (1), (2) și (3) ⇒m(∡B)=2⋅m(∡C)\textcolor{red}{\Rightarrow} \boxed{m(\measuredangle B) = 2 \cdot m(\measuredangle C)}⇒m(∡B)=2⋅m(∡C)
