Question Number 209883 by Ismoiljon_008 last updated on 24/Jul/24
![](https://www.tinkutara.com/question/34509.png)
Commented by Ismoiljon_008 last updated on 24/Jul/24
![help , please](https://www.tinkutara.com/question/Q209884.png)
$$\:\:\:{help}\:,\:{please} \\ $$
Commented by Frix last updated on 25/Jul/24
![I get x=18 AB=BD=CD=6(√5) BE=12(√2) CE=12](https://www.tinkutara.com/question/Q209890.png)
$$\mathrm{I}\:\mathrm{get}\:{x}=\mathrm{18} \\ $$$${AB}={BD}={CD}=\mathrm{6}\sqrt{\mathrm{5}} \\ $$$${BE}=\mathrm{12}\sqrt{\mathrm{2}} \\ $$$${CE}=\mathrm{12} \\ $$
Answered by mr W last updated on 25/Jul/24
![](https://www.tinkutara.com/question/34514.png)
Commented by mr W last updated on 25/Jul/24
![a=6(√5) x+12=(√5)a=30 ⇒x=18](https://www.tinkutara.com/question/Q209900.png)
$${a}=\mathrm{6}\sqrt{\mathrm{5}} \\ $$$${x}+\mathrm{12}=\sqrt{\mathrm{5}}{a}=\mathrm{30} \\ $$$$\Rightarrow{x}=\mathrm{18} \\ $$
Commented by Ismoiljon_008 last updated on 25/Jul/24
![Sorry, please interpret me that how ∠BAD = 45^o](https://www.tinkutara.com/question/Q209902.png)
$$\:\:\:{Sorry},\:{please}\:{interpret}\:{me}\:{that}\:{how} \\ $$$$\:\:\:\angle{BAD}\:=\:\mathrm{45}^{{o}} \: \\ $$
Commented by mr W last updated on 25/Jul/24
![ABDE are cyclic, see below.](https://www.tinkutara.com/question/Q209904.png)
$${ABDE}\:{are}\:{cyclic},\:{see}\:{below}. \\ $$
Commented by mr W last updated on 25/Jul/24
![](https://www.tinkutara.com/question/34515.png)
Commented by Ismoiljon_008 last updated on 25/Jul/24
![ok, I get it. Thank you very much](https://www.tinkutara.com/question/Q209906.png)
$$\:\:{ok},\:{I}\:{get}\:{it}.\:{Thank}\:{you}\:{very}\:{much} \\ $$