Question

*In ∆ABC, on the extension of the side BC , draw a line segment CD ≅ CA . Draw the segment AD . The line segment CE is the angle bisector of ∠ACB. The line segment CF is the median towards AD in ∆ACD. Prove: CF ⊥ CE

Answer 1

CE ⊥ CF because m∠ECA + m∠FCA = 90°

Step-by-step explanation:

In ∆ABC

• On the extension of the side BC , draw a line segment CD ≅ CA
• Draw the segment AD
• The line segment CE is the angle bisector of ∠ACB
• The line segment CF is the median towards AD in ∆ ACD

We want to prove that CF ⊥ CE

Look to the attached figure

In Δ ABC

∵ CE is the bisector of angle ACB

∴ ∠ACE ≅ ∠BCE

In Δ ACD

∵ CA = CD

∴ Δ ACD is an isosceles triangle

∵ AD is the median towards AD

- In any isosceles triangle the median from a vertex to its opposite

  side bisects this vertex

∴ AD bisects ∠ACD

∴ ∠ACF ≅ ∠DCF

∵ BCD is a straight segment

∵ CE , CA , CF are drawn from point C

∴ m∠BCE + m∠ACE + m∠ACF + m∠DCF = 180°

∵ m∠ACE ≅ m∠BCE

∵ m∠ACF ≅ m∠DCF

- Replace m∠BCE by m∠ACE and m∠DCF by m∠ACF

∴ m∠ACE + m∠ACE + m∠ACF + m∠ACF = 180°

∴ 2 m∠ACE + 2 m∠ACF = 180°

- Divide all terms by 2

∴ m∠ACE + m∠ACF = 90°

∴ EC ⊥ CF

CE ⊥ CF because m∠ECA + m∠FCA = 90°

Learn more:

You can learn more about perpendicular lines in brainly.com/question/11223427

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