Centre of the circle is O, tangent CA at A and tangent DB at B intersects each other at point P. Bisectors of $∠$ CAB and $∠$ DBA intersects at point Q on the circle $∠$ APB = 60 $°$
Prove that $△ PAB ≅ △ QAB$

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Solution

$We know that the radius of a circle is perpendicular to its tangent . ∠ OAP = ∠ OBP = 90 ° ∠ OAC = ∠ OBD = 90 ° Hence, OA ⊥ CP at A and OB ⊥ DP at B . ∠ OAB = ∠ OBA . . . (i) (OA and OB are radii) ∠ OAP = ∠ OBP = 90 ° . . . . (i i) Subtracting eq (i) from eq (ii), we get : ∠ OAP - ∠ OAB = ∠ OBP - ∠ OBA \Rightarrow ∠ BAP = ∠ ABP In ∆ APB, we have : ∠ BAP + ∠ ABP + ∠ APB = 180 ° (Angle sum property) \Rightarrow 2 ∠ BAP = 180 ° - 60 ° \Rightarrow ∠ BAP = 60 °$

Since all angles of triangle APB are equal, it is an equilateral triangle.

$In □ AOBP, we have : ∠ OAP + ∠ APB + ∠ PBO + ∠ AOB = 360 ° (Angle addition property of quadrilateral) ∠ AOB = 360 ° - ∠ OAP - ∠ APB - ∠ PBO = 360 ° - 90 ° - 60 ° - 90 ° = 120 ° We know that the angle subtended by the arc at the centre is twice the angle subtended at any part on the circle . ∠ AQB = \frac{1}{2} ∠ AOB = 60 °$

$∠ OAC = ∠ OBD . . . (ii) Adding eq (i) & eq (ii), we get : ∠ OAC + ∠ OAB = ∠ OBD + ∠ OBA \Rightarrow ∠ BAC = ∠ ABD Dividing both sides by 2, we get : \frac{1}{2} ∠ BAC = \frac{1}{2} ∠ ABD \Rightarrow ∠ QAB = ∠ QBA In ∆ AQB, we have : ∠ AQB + ∠ QAB + ∠ QBA = 180 ° (Angle sum property) \Rightarrow 2 ∠ QAB = 180 ° - 60 ° \Rightarrow ∠ QAB = 60 °$

Since, all angles of triangle AQB are equal, it is an equilateral triangle.

AB is common so all the sides, both the triangles are equal.
Hence, by SSS congruency, $QAB ≅ ∆ PAB$

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Centre of the circle is O, tangent CA at A and tangent DB at B intersects each other at point P. Bisectors of ∠CAB and ∠DBA intersects at point Q on the circle ∠APB = 60° Prove that △PAB≅△QAB

Centre of the circle is O, tangent CA at A and tangent DB at B intersects each other at point P. Bisectors of $∠$ CAB and $∠$ DBA intersects at point Q on the circle $∠$ APB = 60 $°$
Prove that $△ PAB ≅ △ QAB$