In quadrilateral $A C B D, A C=A D$ and $A B$ bisects $\angle A$ (See the given figure). Show that $\triangle A B C \cong \triangle A B D$. What can you say about $B C$ and $B D$ ?
Solution:
In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{ABD}$
$A C=A D$ (Given)
$\angle \mathrm{CAB}=\angle \mathrm{DAB}(\mathrm{AB}$ bisects $\angle \mathrm{A})$
$A B=A B$ (Common)
$\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{ABD}$ (By $S A S$ congruence rule)
$\therefore \mathrm{BC}=\mathrm{BD}(\mathrm{By} \mathrm{CPCT})$
Therefore, $B C$ and $B D$ are of equal lengths.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.