In an isosceles triangle $ABC$, with $AB = AC$, the bisectors of $\angle B = \angle C$ intersect each other at $O$.Join $A$ to $O$. Show that $(i) OB = OC $ and $(ii) AO$ bisects $\angle A$.
Answer
Verified
Hint: Join $A$ to $O$. Apply the given condition to the isosceles triangle and use similarity criterion. According to given data we have 3 conditions, $AB = AC \to (1)$ $OB$ is the bisector of $\angle B$ So,$\angle ABO = \angle OBC = \dfrac{1}{2}\angle B \to (2)$ $OC$ is the bisector of$\angle C$ So,$\angle ACO = \angle OCB = \dfrac{1}{2}\angle C$$ \to (3)$ Case-1 So, here we have to prove $OB = OC$ Proof: Now by using condition (1) we can say that, $AB = AC$ From this condition we say that $ \Rightarrow \angle ACB = \angle ABC$ [ Where we know that Angles opposite to equal sides are equal] $\ \Rightarrow \dfrac{1}{2}\angle ACB = \dfrac{1}{2}\angle ABC \\ \\ \ $ $ \Rightarrow \angle OCB = \angle OBC$ [From (2) and (3)] Hence, $OB = OC$ [Sides opposite to equal angles are equal] Hence proved that $OB = OC$. Case-2 We have to prove that $\angle OAB = \angle OAC$ By using case (1) we know that $OB = OC$ And also from $\Delta ABO$ and$\Delta ACO$, we have $ \Rightarrow AB = AC$ (Given) $ \Rightarrow AO = OA$ (Common) $ \Rightarrow OB = OC$(From (case 1)) $\therefore \Delta ABO \cong \Delta ACO$ (By SSS Congruence rule) $ \Rightarrow \angle OAB = \angle OAC$ (CPCPT Theorem)
Hence we have proved that$\angle OAB = \angle OAC$. CASE - 1 CASE - 2 NOTE: In this problem given construction is mandatory to prove the given statements so join $A$ and $O$ points.
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