Question

(a) In the given figure, prove that: BD = BC.



(b) $\text{In the given figure,}\Delta ABC\text{and}\Delta OBC\text{are both isosceles.}$
Prove that $\Delta AOC$ is congruent to $\Delta AOB$.

[4 MARKS]

Answer 1

Each Proof: 2 Marks

(a) In $\Delta ABC$ and $\Delta ABD$


$AB=AB$ [Common side]

$\angle ABC=\angle ABD={90}^{\circ }$ [Given]

$AC=AD$ [Given]

$\Rightarrow \Delta ABC\cong \Delta ABD$ [RHS congruency criteria]

$\therefore BD=BC$ [Corresponding parts of congruent triangles]


(b)

$\text{In}\Delta AOB\text{and}\Delta AOC,AB=AC\text{(Sides of isosceles}\Delta ABC\text{)}.OB=OC\text{(Sides of isosceles}\Delta OBC\text{)}.AO=AO\text{(Common side)}\therefore \Delta AOB\cong \Delta AOC\text{(by SSS congruence)}$