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Step-5

Title:

Grade: 10-a Lesson: S2-L3

Explanation:

From the figure we can observe that

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D is the mid-point of AC and \$\angle ABD = \angle CBD\$.

Consider \$\triangle ADB\$ and \$\triangle CDB\$.

\begin{align} DB &= DB \tag{Common side} \\ AD &= CD \tag{Given} \\ AB &= CB \tag{Given} \\ \end{align}

So, \begin{align} \triangle ADB \cong \triangle CDB \\ \end{align}

Henced proved that \$\triangleADB\$ \$\cong\$ \$\triangleCDB\$ by the SSS congruence rule.

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Steps	Statment	Solution
1	Given	From the figure D is the mid-point of AC and \$\angle ABD = \angle CBD\$.
2	Consider triangles	\$\triangle ADB and \triangle CDB\$
3	Common side	DB = DB
4	Side of triangles	AD = CD and AB = CB
5	So, Congruency	\$\triangle ADB \cong \triangle CDB\$
6	Prove that	Henced proved that \$\triangleADB\$ \$\cong\$ \$\triangleCDB\$ by the SSS congruence rule.

Steps

Statment

Solution

Given

From the figure D is the mid-point of AC and \$\angle ABD = \angle CBD\$.

Consider triangles

\$\triangle ADB and \triangle CDB\$

Common side

DB = DB

Side of triangles

AD = CD and AB = CB

So, Congruency

\$\triangle ADB \cong \triangle CDB\$

Prove that

Henced proved that \$\triangleADB\$ \$\cong\$ \$\triangleCDB\$ by the SSS congruence rule.