Question

In the adjoining figure, measures of different angles are marked. Can we conclude that $∆ABC\cong ∆ADC$ using ASA congruence condition. Explain.

Answer 1

STEP 1 : Finding the value of angles $\angle BCA$ and $\angle CAD$.

In $∆ABC$,

$\angle BCA+\angle CAB+\angle ABC=180°$ [ By angle sum property ]

$\Rightarrow \angle BCA+50°+100°=180°$

$\Rightarrow \angle BCA=180°-150°=30°$

Similarly, In $∆ADC$,

$\angle DCA+\angle CAD+\angle ADC=180°$ [ By angle sum property ]

$\Rightarrow 30°+\angle CAD+100°=180°$

$\Rightarrow \angle CAD=180°-130°=50°$

STEP 2 : Finding whether $∆ABC\cong ∆ADC$ or not

In $∆ABC$ and $∆ADC$,

$\angle BAC=\angle DAC$ [ Each angle $50°$ ]

$AC=AC$ [ Common side ]

$\angle BCA=\angle DCA$ [ Each angle $30°$ ]

Therefore, by ASA congruence condition $∆ABC\cong ∆ADC$.