Question
In Fig., $OA = OB$ and $OD = OC$. Show that
$i. \triangle AOD \cong \triangle BOC$
$ii. AD \| BC$
$i. \triangle AOD \cong \triangle BOC$
$ii. AD \| BC$
✓
Answer
$i.$ You may observe that in $\triangle A O D$ and $\triangle B O C$,
$OA = OB ($Given$)$
$OD = OC$
Also, since $\angle AOD$ and $\angle BOC$ form a pair of vertically opposite angles, we have $\angle A O D=\angle B O C$
So, $\triangle A O D \cong \triangle B O C ($by the $\text{SAS}$ congruence rule$)$
$ii.$ In congruent triangles $\text{AOD}$ and $\text{BOC}$ , the other corresponding parts are also equal.
So, $\angle O A D=\angle O B C$ and these form a pair of alternate angles for line segments $A D$ and $B C$.
Therefore, $A D\| B C$
$OA = OB ($Given$)$
$OD = OC$
Also, since $\angle AOD$ and $\angle BOC$ form a pair of vertically opposite angles, we have $\angle A O D=\angle B O C$
So, $\triangle A O D \cong \triangle B O C ($by the $\text{SAS}$ congruence rule$)$
$ii.$ In congruent triangles $\text{AOD}$ and $\text{BOC}$ , the other corresponding parts are also equal.
So, $\angle O A D=\angle O B C$ and these form a pair of alternate angles for line segments $A D$ and $B C$.
Therefore, $A D\| B C$
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