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Eliseo Soto · Answer 1 · 2022-10-18T16:43:27+0000

$\: \: \: \: \: \: \red{ \underline{ \large{ \pink{ \tt{꧁ A \: N \: S \: W \: E \: R ꧂}}} }}$

✧
$\underline{ \underline{ \large{ \red{ \tt{G \: I\: V \: E\: N}}}} }:$

AC bisects
$\tt{ \angle BAD\: \& \: \angle \: BCD}$

❀
$\underline{ \underline{ \large{ \tt{ \purple{T \: O \: \: F \: I\: N\: D}}}}} :$

Congruence theorem which justifies ∆ ABC
$\cong$ ∆ ADC.

♕
$\underline{ \underline{ \large{ \tt{ \orange{P \: R\:O \: O \: F \: S}}} }}:$

$\begin{array} \hline \text{SN}& \text{Statement} & \text{Reason}\\ \hline \\ \text{1 \: i.}& \sf{In \triangle} \:ABC \: \& \: \triangle \: ACD \: \ \\ & \tt{ \angle ACB= \angle \: ACD(A)}\ & \sf{AC \: bisects \angle \: BCD(Given)} \\ \text{ii} \: & \tt{AC= CA(S)}& \sf{Common \: side} \\ \text{iii} \: & \tt{ \angle \: BAC = \angle \: CAD(A)}& \sf{AC \: bisects \: \angle \: BAD ( Given )}& \\ \\ \hline 2& \tt{ \triangle} \: ABC \cong \triangle{ADC}& \red{\underline{\sf{ \bold{By \: ASA\: axiom}}}} \\ \hline\end{array}$

Hence , We can conclude :

ASA congruence theorem justifies ∆ ABC
$\cong$ ∆ ADC

ツ Hope I helped! ♡

♪ Have a wonderful day / night ! ✎

✎
$\underbrace{ \overbrace{ \mathfrak{Carry \: On \:Learning}}} ☃$

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