MHB Proving the Similarity of Two Acute Triangles with Perpendicular Lines

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Geometry
AI Thread Summary
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.
Albert1
Messages
1,221
Reaction score
0
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 $\triangle ABC$)
 
Mathematics news on Phys.org
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

View attachment 6558
 

Attachments

  • ABC similar to AEF.jpg
    ABC similar to AEF.jpg
    37.1 KB · Views: 112
Last edited by a moderator:
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Replies
1
Views
1K
Replies
5
Views
2K
Replies
3
Views
2K
Replies
1
Views
1K
Replies
2
Views
2K
Replies
4
Views
1K
Replies
11
Views
6K
Back
Top