Question

BCD=ADC
ACB=BDA

To prove:-
AD=BC​

Answer 1

${\tt{\huge{\blue{Solution}}}}$

If ∠BCD = ∠ADC --I

and ∠ACB = ∠BDA---2

${\tt{\red{To\:Proof}}}$

AD = BC

${\tt{\orange{Proof\::}}}$

→In ΔACD and ΔBDC

CD = CD ---------------(Common)

∠ADC=∠BCD ---------(Given)

∠ACB+∠BCD=∠BDA+∠ADC --(FROM eq: 1 & 2)

∴ By SAS criteria ΔACD≈ΔBDC

→Thus, AD=BC -------by C.P.C.T

Hence Proved

Answer 2

Answer:

Solution:-

Let AD and BC intersects at P

Now,∆ APC and ∆ BPD

Angle ACP = Angle BDP (given)

Angle BPD =Angle APC ( Vertically opposite angles)

So, Angle A= Angle B ( Angle of triangle)

In ∆ACD and ∆BDC

Angle C= Angle D (Given)

Angle A = Angle B (Proved above)

CD = CD ( Common )

So,∆ACD =∆ BDC (AAS)

Therefore,AD=BC (CPCT).............[Proved]

Hope it will help you.......