Question

In the figure, ABCD is a trapezium with AD = BC. If ∠ ADC = ∠ DCB, prove that
AC = BD.

Answer 1

Answer:

where is the figure I can't see it

Answer 2

Answer:

(I couldn't see the figure so I assumed it)

Given ABCD is trapezium where AD=BC.

(i) To prove: ∠A=∠B

we can see that AECD is a parallelogram, so sum of adjacent angles =180

o

→∠A+∠E=180

o

→∠A+x=180

o

→∠A=180

o

−x=∠B

Hence proved.

(ii) To prove: ∠C=∠D

sum of adjacent angles in parallelogram is π, so

→∠D∠C+180

o

−2x=180

o

→∠C+∠D=2x

Now

→∠B+∠C=180

o

→180

o

−x+∠C=180

o

=0 ∠C=x, so

∠D=x

And,

∠C=∠D

Hence proved.

(iii) ΔABC=ΔBAD

→ side AB is common.

→AD=BC (given)

so the angle including both the sides is also same,

∠A=∠B. So

ΔABC=ΔBAD (By SAS congruent Rule)

Hence proved.

(iv) As ΔABC=ΔBAD

The third side of both triangles i.e. diagonals are equal AC=BD

DO MARK ME THE BRAINLIEST!!