Question

$△ABD≅△BAC$

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In which $AD=BC$

$∠DAB=∠CBA$

To prove:

(i) $△ABD=△BAC$

(ii) $∠ABD=∠BAC$

Proof: In $△ABD$ and $△ABC$

$AB=AB$ (common)

$∠DAB=∠CBA$ (Given)

$AD=BC$

(i) $△ABD≅△ABC$ ($SAS$ axiom)

(ii) $BD=AC$

(iii) $∠ABD=∠BAC$

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