Question

ABCD is a trapizsem. AB ||DC .AD=BC. prove that angle A = angle B, angle C=angle=D. triangle ABC=triangle BAD. diagonal AC=BC

Answer 1

Hello Mate!

Given : In trap. ABCD, AD = BC.

To Prove : < C = < D, < A = < B, ∆ABC ~ ∆BAD and AC = BC.

To construct : Draw CE || AD.

Proof : AD = BC ( Given ) _(i)

Now, AE || CD and AD || CE so AECD is a ||gm.

AD = CE _(ii)

So from (i) and (ii) we get,

BC = CE.

So, < BEC = < CBE ( <s opp. to equal sides are equal )

or < E = < B

Now, < ABC + < CBE = 180° ( linear pair )

< ABC + < CEB = 180°

< B + < E = 180°

< E = 180° - < B_(iii)

Now, < BAD + < AEC = 180° ( adjacent angle sum is 180° )

< A + < E = 180°

< E = 180° - < A _(iv)

From (iii) and (iv)

180° - < B = 180° - < A

< A = < B

Now, join diagonals.

In ∆BAD and ∆ABC

AB = AB ( Common )

< A = < B ( Proved above )

AD = BC ( Given )

Hence, ∆BAD ~ ∆ABC by SAS congruency.

AC = BD ( c.p.c.t )

In ∆ADC and ∆BCD

AD = BC

CD = CD

AC = BD

Hence ∆ADC ~ ∆BCD by SSS congruency.

< C = < D ( c.p.c.t )

Hence Proved all above part. Q.E.D.

Have great future ahead!