[tex] \: \: \: \: \: \: \red{ \underline{ \large{ \pink{ \tt{꧁ A \: N \: S \: W \: E \: R ꧂}}} }}[/tex]
✧ [tex] \underline{ \underline{ \large{ \red{ \tt{G \: I\: V \: E\: N}}}} }: [/tex]
- AC bisects [tex] \tt{ \angle BAD\: \& \: \angle \: BCD}[/tex]
❀ [tex] \underline{ \underline{ \large{ \tt{ \purple{T \: O \: \: F \: I\: N\: D}}}}} : [/tex]
- Congruence theorem which justifies ∆ ABC [tex] \cong[/tex] ∆ ADC.
♕ [tex] \underline{ \underline{ \large{ \tt{ \orange{P \: R\:O \: O \: F \: S}}} }}: [/tex]
[tex]\begin{array}{ |c| c | c | } \hline \text{SN}& \text{Statement} & \text{Reason}\\ \hline \\ \text{1 \: i.}& \sf{In \triangle} \:ABC \: \& \: \triangle \: ACD \: \ \\ & \tt{ \angle ACB= \angle \: ACD(A)}\ & \sf{AC \: bisects \angle \: BCD(Given)} \\ \text{ii} \: & \tt{AC= CA(S)}& \sf{Common \: side} \\ \text{iii} \: & \tt{ \angle \: BAC = \angle \: CAD(A)}& \sf{AC \: bisects \: \angle \: BAD ( Given )}& \\ \\ \hline 2& \tt{ \triangle} \: ABC \cong \triangle{ADC}& \red{\underline{\sf{ \bold{By \: ASA\: axiom}}}} \\ \hline\end{array}[/tex]
Hence , We can conclude :
- ASA congruence theorem justifies ∆ ABC [tex] \cong[/tex] ∆ ADC
ツ Hope I helped! ♡
♪ Have a wonderful day / night ! ✎
✎ [tex] \underbrace{ \overbrace{ \mathfrak{Carry \: On \:Learning}}} ☃ [/tex]
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