Question

In the given figure, AD = BC and BD = CA, Prove that < ADB = < BCA and < DAB = < CBA

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Answer 1

Answer:

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Answer 2

Answer:

In triangles ABD and BAC, we have

AD=BC (given);

AB=AB (common);

AC=BD (given).

We can use SSS condition to conclude that △ABD≅△BAC. From this we conclude that

∠ADB=∠BCA and ∠DAB=∠CBA.