In the figure abcd is a quadrilateral in which AD=BC and AB parallel CD. prove that the points A B C and D lie on a circle
Answers
Step-by-step explanation:
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.
Given ABCD is a quadrilateral in which AD = BC and ∠ADC = ∠BCD.
To prove ABCD is a cyclic quadrilateral.
Construction Draw DE perpendicular to AB and CF perpendicular to AB.
image
Proof In ∆AED and ∆BFC,
∠ AED = ∠BFC=90° [∵ DE ⊥ AB and CE ⊥ AB]
ED = FC
[ ∵ distance between two parallel lines are equal ]
and AD =BC [given]
image
Then, ∠A = ∠B [by CPCT] … ,(i)
and ∠ADE = ∠BCF [by CPCT]
Since, ABCD is a quadrilateral.
∠A+∠B+∠C+∠D=360°
[∵ sum of all angles of a quadrilateral is 360°]
⇒∠B + ∠B + ∠D + ∠D = 360° [from Eq. (i)]
⇒2 ∠B + 2 ∠D=360°
⇒∠B + ∠D=180°
Hence, ABCD is a cyclic quadrilateral because the sum of opposite angles is 180°.