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Title:

Grade: 10-a Lesson: S2-L3

Explanation:

From the figure we can observe that

$2$

\begin{align} AB = BD \tag{Given}\\ AC = DC \tag{Given}\\ \end{align} and BC is the angle bisector of \$\angle B\$ and \$\angle C\$

Consider \$\triangle ABC\$ and \$\triangle DBC\$.

\begin{align} BC &= BC \tag{Common side} \\ AB &= DB \tag{Given} \\ AC &= DC \tag{Given} \\ \end{align}

So, \begin{align} \triangle ABC \cong \triangle DBC \\ \end{align}

Henced proved that \$\triangleABC\$ \$\cong\$ \$\triangleDBC\$ by the SSS congruence rule.

From the figure we can observe that
$2$

Steps	Statment	Solution
1	Given	From the figure we can observe that
2	Side of triangles	AB = BD and AC = DC
3	bisector	BC is the angle bisector of \$\angle B\$ and \$\angle C\$
4	Consider triangles	\$\triangle ABC and \triangle DBC\$
5	Common side	BC = BC
6	Side of triangles	AB = DB and AC = DC
7	So, Congruency	\$\triangle ABC \cong \triangle DBC\$
8	Prove that	Henced proved that \$\triangleABC\$ \$\cong\$ \$\triangleDBC\$ by the SSS congruence rule.

Steps

Statment

Solution

Given

From the figure we can observe that

Side of triangles

AB = BD and AC = DC

bisector

BC is the angle bisector of \$\angle B\$ and \$\angle C\$

Consider triangles

\$\triangle ABC and \triangle DBC\$

Common side

BC = BC

Side of triangles

AB = DB and AC = DC

So, Congruency

\$\triangle ABC \cong \triangle DBC\$

Prove that

Henced proved that \$\triangleABC\$ \$\cong\$ \$\triangleDBC\$ by the SSS congruence rule.