Seg AB is a diameter of a circle with centre P. Seg AC is
a chord. A secant through P and parallel to seg AC intersects
the tangent drawn at C in D. Prove that line DB is a tangent
to the circle.
Answers
The line DB is proved to be a tangent to the given circle.
Step-by-step explanation:
It is given that,
Segment AB is a diameter of the circle with centre P
Segment AC is a chord
CD is a tangent drawn at point C.
We know that a line is a tangent to a circle if the line is perpendicular to the radius drawn to the point of tangency.
∴ CP is perpendicular to CD
⇒ ∠PCD = 90° ….. (i)
Considering ∆APC, we have
AP = CP ….. [radius of the circle]
∴ ∠PAC = ∠PCA ….. [angles opposite to equal sides are also equal] …. (ii)
Also, we have AC // PD
So,
∠PCA = ∠CPD ……. [alternate angles] ….. (iii)
∠PAC = ∠DPB …… [corresponding angles] …… (iv)
From (i), (ii) & (iii), we get
∠CPD = ∠DPB ….. (v)
Now, in ∆ CPD and ∆ BPD, we have
PD = PD ……. [common side of both the triangle]
∠CPD = ∠DPB …… [from (v)]
CP = BP …… [radius of the circle]
∴ By SAS, ∆ CPD ≅ ∆ BPD
∴ ∠PCD = ∠PBD …… [By Corresponding Parts of Congruent Triangles] …… (vi)
From (i) & (vi), we get
∠PBD = 90°
PB is perpendicular to DB
So, according to the converse tangent theorem if a line is perpendicular to the radius at its endpoint then the line is tangent to the circle.
DB is tangent to the circle at point B.
Hence proved
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