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Lesson Example Discussion Quiz: Class Homework

Quiz Discussion

Title: Congruency of triangles (SSS)

Grade: 10-a Lesson: S2-L3

Explanation:

Quiz: Discussion in Class

Problem Id	Question
Steps 1	\$AB ∥ DC\$, \$AD ∥ BC\$ and \$AC\$ is the diagonal \$ABCD\$. Prove \$\triangleABC \cong \triangleCDA\$. $1$
Steps 2	BC is angle bisector of \$\angleABD\$ and \$\angleACD\$. Prove \$\triangleABC \cong \triangleDBC\$. $2$
Steps 3	\$PA = PB\$, \$QA = QB\$ and \$PQ \bot AB\$, then show that \$\triangle PAQ \cong \triangle PBQ\$. $3$
Steps 4	PQ and RS are the tangents of the circle with center S and SQ is angle bisector of \$\angle S\$ and \$\angleQ\$. Show that \$\triangle PQS \cong \triangle RQS\$. $4$
Steps 5	\$\triangle ABC\$ is a right angled triangle, \$\angleB = 90^circ\$ and DB is the angle bisector of \$\angle B\$. Show that \$\triangle ADB \cong \triangle CDB\$. $5$

Problem Id

Question

Steps 1

\$AB ∥ DC\$, \$AD ∥ BC\$ and \$AC\$ is the diagonal \$ABCD\$. Prove \$\triangleABC \cong \triangleCDA\$.

$1$

Steps 2

BC is angle bisector of \$\angleABD\$ and \$\angleACD\$. Prove \$\triangleABC \cong \triangleDBC\$.

$2$

Steps 3

\$PA = PB\$, \$QA = QB\$ and \$PQ \bot AB\$, then show that \$\triangle PAQ \cong \triangle PBQ\$.

$3$

Steps 4

PQ and RS are the tangents of the circle with center S and SQ is angle bisector of \$\angle S\$ and \$\angleQ\$. Show that \$\triangle PQS \cong \triangle RQS\$.

$4$

Steps 5

\$\triangle ABC\$ is a right angled triangle, \$\angleB = 90^circ\$ and DB is the angle bisector of \$\angle B\$. Show that \$\triangle ADB \cong \triangle CDB\$.

$5$