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

Commented by infinityaction last updated on 11/Apr/22

please solve this problem

$${please}\:{solve}\:{this}\:{problem} \\ $$

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

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

∠BAT=∠BTE=∠TEF=∠TAC=α ⇒AT is bisector of ∠A ⇒((BT)/(TC))=((BA)/(CA))=(9/(11)) ⇒BT=((9×BC)/(9+11))=((9×15)/(20))=((27)/4) ∠EBD=∠EDB=∠TBD=β ⇒BK is bisector of ∠B ⇒((AK)/(KT))=((AB)/(BT))=(9/(27/4))=(4/3) ✓

$$\angle{BAT}=\angle{BTE}=\angle{TEF}=\angle{TAC}=\alpha \\ $$$$\Rightarrow{AT}\:{is}\:{bisector}\:{of}\:\angle{A} \\ $$$$\Rightarrow\frac{{BT}}{{TC}}=\frac{{BA}}{{CA}}=\frac{\mathrm{9}}{\mathrm{11}} \\ $$$$\Rightarrow{BT}=\frac{\mathrm{9}×{BC}}{\mathrm{9}+\mathrm{11}}=\frac{\mathrm{9}×\mathrm{15}}{\mathrm{20}}=\frac{\mathrm{27}}{\mathrm{4}} \\ $$$$\angle{EBD}=\angle{EDB}=\angle{TBD}=\beta \\ $$$$\Rightarrow{BK}\:{is}\:{bisector}\:{of}\:\angle{B} \\ $$$$\Rightarrow\frac{{AK}}{{KT}}=\frac{{AB}}{{BT}}=\frac{\mathrm{9}}{\mathrm{27}/\mathrm{4}}=\frac{\mathrm{4}}{\mathrm{3}}\:\checkmark \\ $$

Commented by Tawa11 last updated on 11/Apr/22

Great sir

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

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