Question

- Question in above attachment solve it!!!



​

Answer 1

$\\ \huge\mathfrak {\underline \blue{ \: \: \: \: \: \: \: \: given \mapsto \: \: \: \: \: \: \: \: }}$

$\\ \large\begin{gathered} \bf \green{in \: a \: traingle} \begin{cases} & \sf{AB \longrightarrow 6} \\ & \sf{ AC \longrightarrow 5} \\ & \sf{BD \longrightarrow 3} \end{cases}\\ \\\end{gathered}$

$\\ \huge\mathfrak {\underline \orange{ \: \: \: \: \: \: \: \: to \: find\mapsto \: \: \: \: \: \: \: \: }}$

$\large\begin{gathered}\bf \pink{in \: a \: traingle} \begin{cases} & \sf{} \\ & \sf{DC \longrightarrow ?} \\ & \sf{ } \end{cases}\\ \\\end{gathered}$

$\\ \huge \mathfrak { \underbrace\purple{diagram}}$

$\large {\red{\boxed { \tt {\pink see \: \: i n \:the \: attachment \: }}}}$

$\\ \huge\mathfrak {\underline \purple{ \: \: \: \: \: \: \: \: solution \mapsto \: \: \: \: \: \: \: \: }}$

As the query said that ∠BAC is bisected by AD and

AD touches BC at D. We have to have DC.

$\large \tt\blue➦ \frac{AB}{BD} = \frac{AC}{DC} \\ \\ \large \sf(by \: cross \: multiplication⇩)\\ \\ \large \tt\blue➦ AB \times DC = AC \times \: B D \\ \\ \large \sf(putting \: the \: values⇩) \\ \\ \large \tt\blue➦ 6 \times DC = 5 \times 3 \\ \\ \large \tt\blue➦ 6 \times DC = 15 \\ \\ \large \tt\blue➦ DC = \frac{15}{6} \\ \\ \large \sf(by \: dividing \frac{15}{6} we \: get) \\ \\ \large \tt\blue➦ DC = 2.5$