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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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