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Question Number 168471 by infinityaction last updated on 11/Apr/22

Commented by infinityaction last updated on 11/Apr/22

$$pleasesolvethisproblem$$

Answered by mr W last updated on 11/Apr/22

Commented by mr W last updated on 11/Apr/22

$$\mathrm{\angle }BAT=\mathrm{\angle }BTE=\mathrm{\angle }TEF=\mathrm{\angle }TAC=\alpha$$$$\Rightarrow ATisbisectorof\mathrm{\angle }A$$$$\Rightarrow \frac{BT}{TC}=\frac{BA}{CA}=\frac{9}{11}$$$$\Rightarrow BT=\frac{9\times BC}{9+11}=\frac{9\times 15}{20}=\frac{27}{4}$$$$\mathrm{\angle }EBD=\mathrm{\angle }EDB=\mathrm{\angle }TBD=\beta$$$$\Rightarrow BKisbisectorof\mathrm{\angle }B$$$$\Rightarrow \frac{AK}{KT}=\frac{AB}{BT}=\frac{9}{27/4}=\frac{4}{3}✓$$

Commented by Tawa11 last updated on 11/Apr/22

$$\mathrm{Great}\mathrm{sir}$$

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