Prove that Delta ABD cong Delta ACD

Question

Prove that  Delta ABD cong Delta ACD

Toppr Admin · Accepted Answer

From the figure-160-160- BD - CD -160-In - -Delta ABD - and - -Delta ACD -160- BD - CD -160- AD - AD - is common-160-angle ADB-angle ADC-90-o- -Delta ABD -cong-Delta ACD - -SAS Axiom-160-Therefore - it is proved-

Prove that
$$ \Delta ABD \cong \Delta ACD $$

From the figure
$$ BD = CD $$
In $$ \Delta ABD $$ and $$ \Delta ACD $$
$$ BD = CD $$
$$ AD = AD $$ is common
$$\angle ADB=\angle ADC=90^o$$
$$ \Delta ABD \cong\Delta ACD $$ (SAS Axiom)
Therefore , it is proved.

Prove that $$ \Delta ABD \cong \Delta ACD $$

From the figure $$ BD = CD $$ In $$ \Delta ABD $$ and $$ \Delta ACD $$ $$ BD = CD $$ $$ AD = AD $$ is common $$\angle ADB=\angle ADC=90^o$$$$ \Delta ABD \cong\Delta ACD $$ (SAS Axiom) Therefore , it is proved.

Prove that
$$ \Delta ABD \cong \Delta ACD $$

From the figure
$$ BD = CD $$
In $$ \Delta ABD $$ and $$ \Delta ACD $$
$$ BD = CD $$
$$ AD = AD $$ is common
$$\angle ADB=\angle ADC=90^o$$
$$ \Delta ABD \cong\Delta ACD $$ (SAS Axiom)
Therefore , it is proved.