SaiBook.us Geometry

Home Course Section

Register Enroll Refer

Step-1

Title:

Grade: 10-a Lesson: S2-L1

Explanation:

From \$\triangleAOD\$ and \$\triangleCOB\$

We can observe that

$1$
\begin{align} AO = BO \\ OD = OC \\ \end{align}

Also angles, \$\angle AOD\$ and \$\angle BOC\$.

Since angles, \$\angle AOD\$ and \$\angle BOC\$ form a pair of vertically opposite angles, we have

\begin{align} \angle AOD = \angle BOC \\ \end{align}

So, \begin{align} \triangle AOD \cong \triangle BOC \\ \end{align}

Henced proved that \$\triangleAOD\$ and \$\cong\$ \$\triangleBOC\$ by the SAS congruence rule.

From \$\triangleAOD\$ and \$\triangleCOB\$
We can observe that
$1$

Steps	Statment	Solution
1	Given	From \$\triangleAOD and \triangleCOB\$ we can observe that
2	Side of triangles	AO = BO and OD = OC
3	Angles	\$\angleAOD and \angleBOC\$
4		Since angles, \$angleAOD and angleBOC\$ from a pair vertically opposite angles we have
5	So	\$\angleAOD and \angleBOC\$
6	Congruency	\$\triangle AOD \cong \triangle BOC\$
7	Prove that	Henced prove that \$\triangleAOD\$ and \$\cong\$ \$\triangleBOC\$ by the SAS congruence rule.

Steps

Statment

Solution

Given

From \$\triangleAOD and \triangleCOB\$ we can observe that

Side of triangles

AO = BO and OD = OC

Angles

\$\angleAOD and \angleBOC\$

Since angles, \$angleAOD and angleBOC\$ from a pair vertically opposite angles we have

\$\angleAOD and \angleBOC\$

Congruency

\$\triangle AOD \cong \triangle BOC\$

Prove that

Henced prove that \$\triangleAOD\$ and \$\cong\$ \$\triangleBOC\$ by the SAS congruence rule.