A tangent PT is drawn parallel to a chord AB of a circle. Prove that APB is an isosceles triangle.
Asked by | 11th Mar, 2013, 12:52: AM
Construction: Join PO and produce to D.
Now, OP is perpendicular TP (tangent makes a 90 degree angle with the radius of the circle at the point of contact)
Also, TP is parallel to AB
∴∠ADP=90° (corresponding angles)
So, OD is perpendicular to AB. Now since, a perpendicular drawn from the center of the circle
to a chord bisects it.
Hence, PD is a bisector of AB. i.e. AD = DB
Now in triangle ADP and BDP
AB =DB (proved above)
∠ADP=∠BDP (both are 90°)
PD = DP (common)
⇒ΔADPΔBDP ( by SAS)
Hence, ∠PAD = ∠PBD (By CPCT)
Thus, APB is an isosceles triangle.
Answered by | 11th Mar, 2013, 04:13: AM
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