Please solve question no.13
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given: P is a point on the chord BC. such that AB = AP
TPT: CP = CQ
proof:
in the triangle ABP, AB = AP
therefore ∠ABP = ∠APB........(1)
since angle subtended by the chord at the circumference are equal.
∠ABC = ∠AQC [angles subtended by the chord AC at circumference ]
∠ABP = ∠AQC ..........(2) [∠ABC = ∠ABP same angle]
∠APB = ∠CPQ ......(3)
from (1), (2) and (3):
∠CPQ = ∠AQC i.e. ∠CPQ = ∠PQC
therefore CP = CQ [sides opposite to equal angles are equal]
hope this helps you.
TPT: CP = CQ
proof:
in the triangle ABP, AB = AP
therefore ∠ABP = ∠APB........(1)
since angle subtended by the chord at the circumference are equal.
∠ABC = ∠AQC [angles subtended by the chord AC at circumference ]
∠ABP = ∠AQC ..........(2) [∠ABC = ∠ABP same angle]
∠APB = ∠CPQ ......(3)
from (1), (2) and (3):
∠CPQ = ∠AQC i.e. ∠CPQ = ∠PQC
therefore CP = CQ [sides opposite to equal angles are equal]
hope this helps you.
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