Mathematics - Properties of Triangle Question with Solution | TestHub

$\angle ABC=\angle AEC=90^{\circ }\Rightarrow △ADB$ and $△AEC$ are similar $\frac{\mathrm{AD}}{\mathrm{AE}}=\frac{\mathrm{BD}}{\mathrm{CE}}=\frac{\mathrm{AB}}{\mathrm{AC}}$

But $BC=25,BD=20,BE=7\Rightarrow DC=15,CE=24$

$$\Rightarrow \frac{\mathrm{AD}}{\mathrm{AE}}=\frac{20}{24}=\frac{5}{6}$$

$\frac{AB}{BC}=\frac{AE+EB}{AD+DC}=\frac{AE+7}{AD+15}=\frac{5}{6}$

$\Rightarrow AD=15,AE=18$

$\Rightarrow D$ is mid points hypotenuous AC of $△AEC$

$$\Rightarrow \mathrm{DE}=\frac{1}{2}\mathrm{AC}=15$$

$\angle DEF=90^{\circ }\Rightarrow AP=\frac{1}{2}AE=9(△ADE$ is isosceles $)$ as $AD=DC=DE$ $\Rightarrow G,F,E$ and D are concyclic and $\mathrm{D},\mathrm{E},\mathrm{B}$ and C are concyclic too. (Ptolmey's Th.) $\Rightarrow \angle AFG=\angle AADE=\angle ABC=(\pi -2A)\Rightarrow GF\Vert CB$ $\frac{\mathrm{AK}}{\mathrm{AP}}=\frac{\mathrm{AP}}{\mathrm{AB}}\Rightarrow \mathrm{AK}=\frac{\mathrm{AF}\cdot \mathrm{AP}}{\mathrm{AB}}=\frac{9.24}{25}\Rightarrow AP=CE=24$