the adjoining figure ABCD is a
quadrilateral
angle
proved that point ABC and
in whith A B = BC
- BCD
these
are
Consillic.
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It is given that ABCD is a quadrilateral in which AD=BC and ∠ADC=∠BCD
Construct DE⊥AB and CF⊥AB
Consider △ADE and △BCF
We know that
∠AED+∠BFC=90
o
From the figure it can be written as
∠ADE=∠ADC−90
o
=∠BCD−90
o
=∠BCF
It is given that
AD=BC
By AAS congruence criterion
△ADE≃△BCF
∠A=∠B (c.p.c.t)
We know that the sum of all the angles of a quadrilateral is 360
o
∠A+∠B+∠C+∠D=360
o
By substituting the values
2∠B+2∠D=360
o
By taking 2 as common
2(∠B+∠D)=360
o
By division
∠B+∠D=180
o
So, ABCD is a cyclic quadrilateral.
Therefore, it is proved that the points A,B,C and D lie on a circle.
Step-by-step explanation:
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