Prove that the opposite sides of a quadrilateral circumscribe a circle subtend supplementary angles at the centre of the circle
Answers
Answer:
Step-by-step explanation:
consider a quad ABCD which circumscribes a circle with centre O.
the points where the circles touches the quad name them PQRS. tangents to the circle will be formed. now join the centre O to the points ABCD and PQRS. many triangles will be formed. you can prove these adjacent triangles congruent by SSS congruency. by CPCT the angles will be equal. when you add all the angles they will form 360. by calculation you will get your answer.
Answer:
Step-by-step explanation:
To prove ...
Angle aob + angle doc = 180 °
Proof ....
In triangle rob and boq ....
Rb = bq {pair of tangent }
Ro = oq ( radius of same circle )
Bo = bo ( common )
By sss ... triangle rob is congruent to triangle boq
By cpct .......
angle 1 = angle 2 ......[1]
Similarly ,
Triangle qoc is congruent to cop.... by cpct..
Angle 3 = angle 4 ........... [2]
Triangle pod is congruent to dos .... by cpct ..
Angle 5 = angle 6 ...............[3]
Triangle soa is congruent to aor and by cpct ....
Angle 7 = angle 8 ...............[4]
NOW,
By angle sum property ....
Angle 1 + angle2 + angle3+ angle4 +5 +6+7+8 = 360 °
Angle 1+ angle 1 + angle 4 + angle4 + angle 5+angle5 +angle 8 + angle 8 = 360 ° ...... { by eqn [1],[2],[3] and [4] }
2(angle 1 + 8 + angle 4 + 5 ) = 360
{Now we know angle 1 + 8 is angle aob and angle 4+5 is doc by diagram}
so angle 1+8 + 4+5 = 180 °
Therefore ...
Angle aob + doc = 180 °
HENCE PROVED ......
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