In Δ ABC, the perpendicular from vertices A and B on their opposite sides meet (when produced) the circumcircle of the triangle at points D and E respectively. Prove that: arc CD = arc CE
Given: In ΔABC, the perpendiculars from vertices A and B on their opposite sides meet (when produced) the circumcircle of the triangle at points D and E respectively.
To prove: Arc CD = Arc CE
Construction: Join CE and CD
In ΔAPM and ∠BPL
∠AMP = ∠BLP [each = 90°]
∠1 = ∠2 [vertically opposite angles]
ΔAPM ~ ΔBPL [AA postulate]
∴ Third angle = Third angle
∴ ∠3 = ∠4
∴ Arc which subtends equal angle at the
Circumference of the circle are also equal.
∴ Arc CD = Arc CE
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