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Question Number 179453 by cherokeesay last updated on 29/Oct/22

Answered by mr W last updated on 29/Oct/22

Commented by mr W last updated on 29/Oct/22

$$\mathrm{\Delta }AFD\sim \mathrm{\Delta }ABC$$$$\Rightarrow \frac{DF}{AF}=\frac{CB}{AB}$$$$\mathrm{\angle }AEM=\mathrm{\angle }ABM=90°$$$$\Rightarrow AEMBiscyclic.$$$$\Rightarrow \mathrm{\angle }AEF=\mathrm{\angle }AMB$$$$\Rightarrow \mathrm{\Delta }AEF\sim \mathrm{\Delta }AMB$$$$\frac{EF}{AF}=\frac{BM}{AB}=\frac{1}{2}\times \frac{CB}{AB}=\frac{1}{2}\times \frac{DF}{AF}$$$$\Rightarrow EF=\frac{DF}{2}$$$$i.e.EismidpointofDF.$$$$\Rightarrow DE=EF✓$$

Commented by cherokeesay last updated on 29/Oct/22

$$thankyousomuchmaster.$$

Commented by Tawa11 last updated on 30/Oct/22

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

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