Question

in qudrilateral ABCD AB=CD BC=AD S.T ​

Answer 1

Answer:

Step-by-step explanation:

Given that :-

Quadrilateral ABCD . in which

AB = CD

BC = AD .

Show that :-

★ ∆ABC ≅ ∆CDA

Construction :-

★ join A to C .

Proof :-

In triangle ABC and CDA

AB = CD (given)

BC = AD (given)

AC = AC (common)

therefore,

∆ABC ≅ ∆CDA

(by SSS criteria)

hence proved ✓✓

criteria for congruence of triangle :-

SSS :- side, side ,side :- two triangles are congruence when all side of one triangle is equal to the another triangle.

AAA :- angle, angle, angle :- two triangles are congruence when all angles of one triangle is equal to the another triangle.

SAS :- side, angle ,side :- two triangles are congruence when two sides and angle between them are equal to another triangle .

AAS :- angle, angle, side :- two triangles are congruence when two angles and the side, at which both angles are situated, are equal to the another triangle.

RHS :- right angle , hypotenuse, side :- two triangles are congruence when a triangle is right angled . and their hypotenuse and anyone side is equal to the another triangle.

Answer 2

Step-by-step explanation:

In quadrilateral ABCD, we have AB=CD and AD=BC

⇒ Now, join segment AC.

⇒ In △ABC and △ACD, we have

⇒ AB = CD [Given]

⇒ AD = BC [Given]

⇒ AC = AC [Common side]

So, by SSS criteria,

⇒ △ABC≅△ACD

∴ ∠DAC = ∠BCA [By CPCT]

∴ CD∥AB [∵ Alternate angles are equal] ---- ( 1 )

⇒ ∠DCA=∠CAB [By CPCT]

∴ AD∥BC [∵ Alternate angles are equal] --- ( 2 )

From ( 1 ) and ( 2 ) we get that opposite sides of quadrilateral are parallel.

∴ ABCD is a parallelogram.

solution