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Example 3 : AB is a diameter of the circle, CD is a chord equal to theradius of the circle. AC and BD when extended intersect at a point E. Prove that AEB = 60Â°?

Answers (1)

D Deependra Verma

$CD=radius,;r\*OC=OD=radius\* riangle OCD;is;equilateral;triangle.\* Rightarrow angle DCO=angle COD=angle ODC=60^circ\* angle ACB=90^circ;;;;;;;;;;;;;;;;;;;;;;(Angle;in;semicircle)\* angle DOC=2angle DBC;;;;;;;;;;;;;;(half;angle)\* angle DBC=30^circ\* angle ECB+angle BCA=180^circ;;;;;;;;;;;;;(linear;pair)\*angle ECB=180-90=90^circ\* In,; riangle ECB\* angle CEB+angle ECB+angle CBE=180^circ\* angle CEB+90^circ+30^circ=180^circ\* angle CEB=60^circ\*Hence,;proved;that;angle CEB=angle AEB=60^circ$

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