In figure- ADBC is a quadrilateral in which angle ABC - angle ABD and BC - BD then show that Delta ABC cong Delta ABD-

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Toppr Admin · Accepted Answer

In -Delta ABC- and -Delta ABD-angle ABC - -angle ABD- -given-BC - BD- -given-AB - AB- -common side-similarly- -Delta ABC -cong -Delta ABD-By SAS congruency property-Hence proved-

In figure, $$ADBC$$ is a quadrilateral in which $$\angle ABC = \angle ABD$$ and $$BC = BD$$ then show that $$\Delta ABC \cong \Delta ABD$$.

In $$\Delta ABC$$ and $$\Delta ABD$$
$$\angle ABC = \angle ABD$$ (given)
$$BC = BD$$ (given)
$$AB = AB$$ (common side)
similarly, $$\Delta ABC \cong \Delta ABD$$
(By SAS congruency property)
Hence proved.

In figure, $$ADBC$$ is a quadrilateral in which $$\angle ABC = \angle ABD$$ and $$BC = BD$$ then show that $$\Delta ABC \cong \Delta ABD$$.

In $$\Delta ABC$$ and $$\Delta ABD$$$$\angle ABC = \angle ABD$$ (given)$$BC = BD$$ (given)$$AB = AB$$ (common side)similarly, $$\Delta ABC \cong \Delta ABD$$(By SAS congruency property)Hence proved.

In $$\Delta ABC$$ and $$\Delta ABD$$
$$\angle ABC = \angle ABD$$ (given)
$$BC = BD$$ (given)
$$AB = AB$$ (common side)
similarly, $$\Delta ABC \cong \Delta ABD$$
(By SAS congruency property)
Hence proved.