A student constructs the angle bisectors of a triangle and then uses their intersection point as the center of a circle. The stu
2 answers:
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Answer:
Hence proved that m∠AOD = m∠BCD.
Step-by-step explanation:
Given:
A Trapezoid with AC and BD as diagonals with O as intersection point to it.
Four angles formed at A and D are equal.
To Prove:
m∠AOD = m∠BCD
Proof:
Consider a Quadrilateral ABCD as trapezoid with diagonals as AC and BD
(Refer the Attachment).
And AB=CD which means the Quadrilateral ABCD isosceles in nature
i.e. angle A=angle D
Now
Consider triangle AOD and triangle BCD
Apply sum of all angles of triangle is 180.
And also
m∠2+∠AOD+m∠3=180
∠AOD=180-m∠2-m∠3..............equation(1)
And for Triangle BCD
m∠BCD=180-m∠3-m∠4,.....................Equation(2)
∠A=∠D................ as Quadrilateral is Isosceles.
i.e.∠2=∠3
That means by transitive property,
∠1=∠4=∠3=∠2
Using this values in above equation(1) we get ,
∠AOD=180-∠3-∠4.......................(3)
Now comparing equation 2 and 3 we get
RHS is equal for both equation that means LHS is also equal .
Therefore ∠AOD=∠BCD
Hence proved.
