MHB Proving the Similarity of Two Acute Triangles with Perpendicular Lines

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In triangle ABC, with points D, E, and F on sides BC, AC, and AB respectively, the conditions of perpendicularity (AD ⊥ BC, DE ⊥ AC, DF ⊥ AB) lead to the conclusion that triangle ABC is similar to triangle AEF. This similarity is established through the angles formed by the perpendicular lines, which maintain angle congruence. Additionally, it is proven that line AO, where O is the circumcenter of triangle ABC, is perpendicular to line EF. The geometric relationships and properties of acute triangles are crucial in this proof. The findings reinforce the significance of perpendicular lines in establishing triangle similarity and relationships in geometry.
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Acute triangle $ABC$,3 points $D,E,F $ are on $\overline{BC},\overline{AC},\overline{AB}$
respectively ,
if $\overline{AD}\perp \overline{BC} ,\overline{DE}\perp \overline {AC} $ and $\overline{DF}\perp \overline {AB}$
prove :
(1)$\triangle ABC \sim \triangle AEF$
(2) $\overline{AO}\perp \overline {EF} $
(hrere $O$ is the circumcenter of $\triangle ABC$)
 
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Albert said:
Acute triangle $ABC$,3 points $D,E,F $ are on $\overline{BC},\overline{AC},\overline{AB}$
respectively ,
if $\overline{AD}\perp \overline{BC} ,\overline{DE}\perp \overline {AC} $ and $\overline{DF}\perp \overline {AB}$
prove :
(1)$\triangle ABC \sim \triangle AEF$
(2) $\overline{AO}\perp \overline {EF} $
(hrere $O$ is the circumcenter of $\triangle ABC$)
hint:
(1) prove $\angle AEF=\angle B$
(2) Prove $\angle BAO+\angle AFE=90^o$
 
Albert said:
Acute triangle $ABC$,3 points $D,E,F $ are on $\overline{BC},\overline{AC},\overline{AB}$
respectively ,
if $\overline{AD}\perp \overline{BC} ,\overline{DE}\perp \overline {AC} $ and $\overline{DF}\perp \overline {AB}$
prove :
(1)$\triangle ABC \sim \triangle AEF$
(2) $\overline{AO}\perp \overline {EF} $
(hrere $O$ is the circumcenter of $\triangle ABC$)
solution :
The tags of all the related angles are maked with ,hope you can figure them out

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