Question

In a given fig 2.65 , AD divides $\angle$BAC in the ratio 1:3 and AD = DB. Determine the value of x.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{AD}\mathrm{divide}\angle \mathrm{BAC}\hspace{0.17em}\mathrm{into}y\mathrm{and}3y. \\ \mathrm{Then},\angle \mathrm{BAD}=y\mathrm{and}\angle \mathrm{DAC}=3y \\ \mathrm{Also},\angle \mathrm{ABD}=y(\because \mathrm{AD}=\mathrm{DB}) \\ \mathrm{Here},108°+3\mathrm{y}+\mathrm{y}=180°(\because \mathrm{BAE}\mathrm{is}\mathrm{a}\mathrm{straight}\mathrm{line}) \\ \Rightarrow 4y=72° \\ \Rightarrow y=18°...\left(\mathrm{i}\right) \\ \mathrm{In}△\mathrm{ABC},\mathrm{we}\mathrm{have}, \\ 3y+y+y+x=180°\left(\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{measure}\mathrm{of}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\right) \\ \Rightarrow 5y+x=180° \\ \Rightarrow 5\times 18+x=180° \\ \Rightarrow x=90° \end{gathered}$$