Question

plz help to find this ​

Answer 1

Answer:

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

(condition for cyclic quadrilateral)

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:

I hope the answer will help you.

Answer 2

Answer:

your proof is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: prove \: that : \\ points \: A,B,C,D \: lie \: on \: circle \\ ie \: are \: concyclic \\ \\ AD = BC \: \: \: \: \: (given) \\so \: hence \\ \: ∠DAE=∠CBF \\ moreover \: as \: \: A - E - B - F \\ \\ ∠DAB = ∠CBA \: \: \: \: \: \: \: (1) \\ \\ ∠ADC=∠BCD \: \: \: \: \: \: \: \: (2)$

$so \: by \: angle \: sum \: property \: of \\ quadrilateral \\ we \: get \\ \\ ∠ADC + ∠BCD + ∠DAB + ∠CBA = 360 \\ \\ ∠ADC+∠ABC + ∠ADC+∠ABC = 360 \\ \\ 2.∠ADC+2.∠ABC = 360 \\ \\ ∠ADC+∠ABC = 180 \\ \\ by \: converse \: of \: cyclic \: quadrilateral \: \\ theorem \\ we \: conclude \\ \\ □ABCD \: is \: cyclic \\ \\ hence \: then \\ \: points \: A,B,C,D \: must \: lie \: on \: circle \\ so \: that \: circumcircle \: passes \: through \: it$