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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 - ICSE Class 10 - Mathematics

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Question

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

Solution

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
Proof :

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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Solution 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 Concept: Chord Properties - the Perpendicular to a Chord from the Center Bisects the Chord (Without Proof).
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