Question

HELP ME ANYONE MR BRAINLY , OR ANY SMART MATH PEOPLE.

Answer 1

There are two ways ,let's check

Proof:-

$\boxed{\begin{array}{c|c}{\boxed{\bf{Statement}}}&{\boxed{\bf{Reason}}}\\ \sf AB\cong CB &\sf Given \\ \sf AD\cong CD &\sf Given \\ \sf <BAD\cong <BCD &\sf Given \\ \sf BD\cong BD &\sf Common\:side\end{array}}$

So

$\\ \rm\rightarrowtail \triangle ABD\cong \triangle CBD (SSS)\:or\:(SAS)$

Answer 2

Given ↷

• AB ≊ CB
• AD ≊ CD
• ∠BAD ≊ ∠BCD
• BD bisects ∠ABC

To prove ↷

• ∆ABD ≊ ∆CBD

Proof ↷

In ∆ABD & ∆CBD,

Statement 1

• AB ≊ CB

Reason

• Given

Statement 2

• AD ≊ CD

Reason

• Given

Statement 3

• ∠BAD ≊ ∠BCD

Reason

• Given

Statement 4

• BD ≊ BD

Reason

• Common side between∆ABD & ∆CBD

Statement 5

• ∠BDA ≊ ∠BDC

Reason

• Angles formed on the base of bisector are equal

Hence , ∆ABD ≊ ∆CBD by side angle side criterion