If two chords AB and CD of a circle AYDZBWCX intersect at right angles. prove that arc CXA + arc DZB = arc AYD + arc BWC = semicircle
Answers
Answered by
19
Given: AB and CD are two chords of a circle, such that AB⊥ CD
let the chords intersect at O.
TPT: arc CXA + arc DZB = arc AYD + arc CWB = semicircle
construction : join CA , AD , DB and BC
proof:-angle subtended by chord AC is ∠CBA and angle subtended by chord BD is ∠BCD.
in the Δ ,
∠CBA+∠BCD = 180- ∠COB
=180-90=90
since angle subtended in semicircle is 90°
⇒ arc CXA + arc DZB = semi circle
similarly ,
arc AYD + arc CWB = semicircle
let the chords intersect at O.
TPT: arc CXA + arc DZB = arc AYD + arc CWB = semicircle
construction : join CA , AD , DB and BC
proof:-angle subtended by chord AC is ∠CBA and angle subtended by chord BD is ∠BCD.
in the Δ ,
∠CBA+∠BCD = 180- ∠COB
=180-90=90
since angle subtended in semicircle is 90°
⇒ arc CXA + arc DZB = semi circle
similarly ,
arc AYD + arc CWB = semicircle
Answered by
6
Answer:
HOPE IT HELPS
PLS MARK IT AS BRAINLIEST
AND DO FOLLOW ME
Attachments:
![](https://hi-static.z-dn.net/files/dee/f95e412f3772de4ece8e04d91e108b84.jpg)
![](https://hi-static.z-dn.net/files/df3/b90f96727afd831518e163e420409370.jpg)
Similar questions