Question

center of the circle is O,tangent CA at A and tangent DB at B intersect each other at point P . bisector of angle CAB and angle DBA interest at point Q on the circle .angle APB =60° prove that ∆PAB=∆QAB​

Answer 1

In the figure AB || CD ......... (Given)

∴∠ABC=∠BCD .......... (Alternate angles)

But ∠ABC=55

o

......... (Given)

∴∠BCD=55

o

In the figure ∠BOD=2∠BCD ........ (Central angle formed by an arc is twice the inscribed angle formed by the same arc)

∴∠BOD=2×55

o

=110

o

∴∠BOD=110

o

....... (i)

But ∠BOD+∠BPD=180

o

........ [Opposite angles of a quadrilateral are supplementary]

∴∠BPD=180

o

−∠BOD=180

o

−110

o

=70

o

∴∠BPD=70

o

......... (ii)

Answer 2

Answer:

We know that the tangents to a circle from an external point are equal in length so PA= PB. PA =PB ∠PBA = ∠PAB [Angles opposite to the ...