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Question
The altitude from the vertices A,B and C of the triangle ABC meet its circumcircle at D,E and F respectively. The complex numbers representing the points D,E and F are z1,z2 and z3 respectively. If z3−z1z2−z1 is purely real, then the triangle ABC is right angle at
The altitude from the vertices A,B and C of the triangle ABC meet its circumcircle at D,E and F respectively. The complex numbers representing the points D,E and F are z1,z2 and z3 respectively. If z3−z1z2−z1 is purely real, then the triangle ABC is right angle at
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Solution
The correct option is A A
∠FDA=∠FCA=90∘−A
∠ADE=∠ABE=90∘−A
⇒∠FDE=180∘−2A=π−2A
Similarly ∠DFE=π−2C and ∠DEF=π−2B
The angles of ΔDEF are π−2A,π−2B and π−2C, respectively.
Also it is given that z3−z1z2−z1 is purely real. Hence
arg(z3−z1z2−z1)=0 or π
⇒π−2A=0 or π
⇒A=π2 or 0 (not permissible)
Hence triangle ABC is right angled at A.
∠FDA=∠FCA=90∘−A
∠ADE=∠ABE=90∘−A
⇒∠FDE=180∘−2A=π−2A
Similarly ∠DFE=π−2C and ∠DEF=π−2B
The angles of ΔDEF are π−2A,π−2B and π−2C, respectively.
Also it is given that z3−z1z2−z1 is purely real. Hence
arg(z3−z1z2−z1)=0 or π
⇒π−2A=0 or π
⇒A=π2 or 0 (not permissible)
Hence triangle ABC is right angled at A.
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