Question

In the figure, In the figure, prove that ∆ABC ≅ ∆ADC ,if < B =< D, AB = CD , BC = AD and ABCD a parallelogram​

Answer 1

answer: proved ,they are congruent

Explanation:

Given that ABCD is a parallelogram, then AB = CD (opposite sides of a parallelogram). It is given that BC = AD. Therefore, triangles ABC and ADC share two equal sides, AB = CD and BC = AD.

Moreover, if <B = <D, then these triangles also share an equal angle. To see why this is true, we can extend BC and AD to intersect at point E. Since ABCD is a parallelogram, we know that <ADE = <CBE. However, since <B = <D, we know that <CBE = <ABD. Therefore, we have <ADE = <ABD, implying that triangles ABC and ADC share an equal angle.

Therefore, by the Side-Angle-Side (SAS) congruence criterion, triangles ABC and ADC are congruent.

Answer 2

Explanation:

In ∆ ADC and ∆ ABC,

<B = <D(Given)

BC = AD ( given)

AB = CD(Given)

By SAS rule of congruency,

∆ABC ≅ ∆ADC(Proved)