Two triangles ABC and DBC have common base BC- Suppose AB- DC and angle ABC - angle BCD - then prove that AC - BD-

Question

Two triangles ABC and DBC have common base BC- Suppose AB- DC  and angle ABC - angle BCD - then prove that AC - BD-

Toppr Admin · Accepted Answer

In-160-bigtriangleup-160- ABC - and -bigtriangleup DBC -AB - DC - -Given-160-angle-160- ABC - -angle BCD - -Given-160-BC - BC - -common side-160-therefore-160- -bigtriangleup-160- ABC -cong -bigtriangleup-160- DBC - -By -SAS- postulate-160-implies AC - BD - - Corresponding sides of congruent triangles-

Two triangles $$ABC$$ and $$DBC$$ have common base $$BC$$. Suppose $$AB= DC $$ and $$\angle ABC = \angle BCD $$, then prove that $$AC = BD$$.

In $$\bigtriangleup ABC $$ and $$\bigtriangleup DBC $$
$$AB = DC $$ (Given)
$$\angle ABC = \angle BCD $$ (Given)
$$BC = BC $$ (common side)
$$\therefore \bigtriangleup ABC \cong \bigtriangleup DBC $$ (By $$SAS$$ postulate)
$$\implies AC = BD $$ ( Corresponding sides of congruent triangles)

Two triangles $$ABC$$ and $$DBC$$ have common base $$BC$$. Suppose $$AB= DC $$ and $$\angle ABC = \angle BCD $$, then prove that $$AC = BD$$.

In $$\bigtriangleup ABC $$ and $$\bigtriangleup DBC $$$$AB = DC $$ (Given) $$\angle ABC = \angle BCD $$ (Given) $$BC = BC $$ (common side) $$\therefore \bigtriangleup ABC \cong \bigtriangleup DBC $$ (By $$SAS$$ postulate) $$\implies AC = BD $$ ( Corresponding sides of congruent triangles)

In $$\bigtriangleup ABC $$ and $$\bigtriangleup DBC $$
$$AB = DC $$ (Given)
$$\angle ABC = \angle BCD $$ (Given)
$$BC = BC $$ (common side)
$$\therefore \bigtriangleup ABC \cong \bigtriangleup DBC $$ (By $$SAS$$ postulate)
$$\implies AC = BD $$ ( Corresponding sides of congruent triangles)