Question

In the given figure $AB=AD$ and $AC$ bisects $\angle BAD$. Prove that

(i) $∆ABC\cong ∆ADC$

(ii) $AC$ bisects $\angle BCD$.

Answer 1

STEP 1 : Proving that $∆ABC\cong ∆ADC$

In $∆ABC$ and $∆ADC$,

$BA=DA$ [ Given ]

$\angle BAC=\angle DAC$ [ Given ]

$AC=AC$ [ Common Side ]

Therefore, by SAS congruence condition $∆ABC\cong ∆ADC$

STEP 2 : Proving that $AC$ bisects $\angle BCD$

Since, $∆ABC\cong ∆ADC$

$\therefore \angle BCA=\angle DCA$ [ By C.P.C.T. ]

$\Rightarrow \angle BCA=\angle DCA=\frac{1}{2}\angle BCD$

Therefore, we can say that $AC$ bisects $\angle BCD$