2) Given chord AB = chord BC=
chord AC,<BAP -< CAP
To prove: Seg CB=seg Co.
Answers
If chord AB = chord BC = chord AC, <BAP = < CAP, then Segment CB = segment CQ is proved.
Step-by-step explanation:
Since it is given, chord AB = chord BC = chord AC
∴ ∆ ABC will be an equilateral triangle so, ∠CAB = ∠ABC = ∠ ACB = 60°
Also given, ∠BAP = ∠CAP = ½ × ∠CAB = 30°
From the figure attached below, we get
∠ACB + ∠BCQ = 180° …… [Linear pair]
⇒ 60° + ∠BCQ = 180°
⇒ ∠BCQ = 180° - 60°
⇒ ∠BCQ = 120°
We know that an angle formed by the two secants is one half of the difference of its intercepted arcs.
Therefore, we can say
∠AQB = ½ * [arcAB – arcCP] = ½ * [120° - 60°] = ½ * 60° = 30°
Now, consider ∆CBQ, applying angle sum property, we get
∠AQB + ∠CBQ + ∠BCQ = 180°
⇒ 30° + angle CBQ + 120° = 180°
⇒ ∠CBQ = 180° - (120° + 30°)
⇒ ∠CBQ = 30°
Since in ∆CBQ, ∠CBQ = ∠ AQB = 30°
∴ ∆CBQ is an isosceles triangle
⇒ seg CB = seg CQ
Hence proved
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