In a right triangle ABC,right-angled at A,a s | vedclass

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(D) In $\Delta ABC$,$\angle A = 90^\circ$. Let $\angle C = 	heta$. Then $\angle B = 90^\circ - 	heta$.Since $AC$ is the diameter of the semicircle,$

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$\frac{AB \cdot AD}{\sqrt{AB^2 - AD^2}}$

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(D) In $\Delta ABC$,$\angle A = 90^\circ$. Let $\angle C = 	heta$. Then $\angle B = 90^\circ - 	heta$.Since $AC$ is the diameter of the semicircle,$\angle ADC = 90^\circ$ (angle in a semicircle).In $\Delta ADC$,$\cos 	heta = \frac{AD}{AC} \Rightarrow AC = \frac{AD}{\cos 	heta}$.In $\Delta ABC$,$	an 	heta = \frac{AB}{AC} \Rightarrow \cos 	heta = \frac{AC}{\sqrt{AB^2 + AC^2}}$.Substituting $\cos 	heta$ in the first equation: $AC = \frac{AD \cdot \sqrt{AB^2 + AC^2}}{AC}$.$AC^2 = AD \sqrt{AB^2 + AC^2} \Rightarrow AC^4 = AD^2(AB^2 + AC^2)$.$AC^2(AC^2 - AD^2) = AD^2 \cdot AB^2$.$AC^2 = \frac{AD^2 \cdot AB^2}{AC^2 - AD^2}$.Taking the square root,$AC = \frac{AB \cdot AD}{\sqrt{AC^2 - AD^2}}$.Alternatively,using similar triangles $\Delta ADC \sim \Delta BAC$,we have $\frac{AD}{AB} = \frac{AC}{BC} = \frac{DC}{AC}$.From $\Delta ADC$,$AD = AC \cos 	heta$ and $DC = AC \sin 	heta$. From $\Delta ABC$,$AB = AC 	an 	heta$.$AC = \frac{AB \cdot AD}{\sqrt{AB^2 - AD^2}}$.

In a right triangle $ABC$,right-angled at $A$,a semicircle is described on the leg $AC$ as diameter. If $D$ is the point of intersection of the hypotenuse $BC$ and the semicircle,then the length $AC$ is equal to:

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