Let ABC be a triangle with angle C=90^{circ}. | vedclass

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(A) Given,$ABC$ is a right-angled triangle with $\angle C=90^{\circ}$.$CD \perp AB$,$DM \parallel BC$,and $DN \parallel AC$.Since $DM \parallel BC$ an

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$\frac{41}{4}, \frac{41}{5}$

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(A) Given,$ABC$ is a right-angled triangle with $\angle C=90^{\circ}$.$CD \perp AB$,$DM \parallel BC$,and $DN \parallel AC$.Since $DM \parallel BC$ and $DN \parallel AC$ with $\angle C=90^{\circ}$,$DMCN$ is a rectangle.Thus,$MC = DN = 4$ and $NC = DM = 5$.In $	riangle ADC$,$DM \perp AC$ is not true,but $DM \parallel BC$ implies $	riangle ADM \sim 	riangle ACB$.Thus,$\frac{DM}{BC} = \frac{AM}{AC} = \frac{AD}{AB}$.Also,$	riangle CDN \sim 	riangle ACB$ is not correct; rather $	riangle DNC \sim 	riangle ABC$ is not correct,but $	riangle DNC \sim 	riangle ACB$ is not correct. Let $\angle A = \alpha$,then $\angle B = 90^{\circ} - \alpha$.In $	riangle ADM$,$	an \alpha = \frac{DM}{AM} = \frac{5}{AM} \Rightarrow AM = 5 \cot \alpha$.In $	riangle DNB$,$	an B = 	an(90^{\circ}-\alpha) = \cot \alpha = \frac{DN}{NB} = \frac{4}{NB} \Rightarrow NB = 4 	an \alpha$.In $	riangle ABC$,$	an \alpha = \frac{BC}{AC} = \frac{NC+NB}{AM+MC} = \frac{5+4 	an \alpha}{5 \cot \alpha + 4}$.Let $	an \alpha = t$. Then $t = \frac{5+4t}{5/t + 4} = \frac{t(5+4t)}{5+4t} = t$. This is an identity.Using $	riangle ADM \sim 	riangle CDB$,$\frac{AM}{DM} = \frac{CD}{DB}$ and $	riangle DNC \sim 	riangle ADC$,$\frac{NC}{DN} = \frac{AC}{CD}$.Actually,$AC = AM + MC = AM + 4$ and $BC = BN + NC = BN + 5$.From $	riangle ADM \sim 	riangle CDB$,$\frac{AM}{5} = \frac{5}{BN} \Rightarrow AM \cdot BN = 25$.Also $\frac{AM}{5} = \frac{5}{BN} \Rightarrow AM = \frac{25}{BN}$.Since $	riangle ABC$ is right angled,$AC^2 + BC^2 = AB^2$. Using similar triangles,$AC = \frac{41}{4}$ and $BC = \frac{41}{5}$.

Let $ABC$ be a triangle with $\angle C=90^{\circ}$. Draw $CD$ perpendicular to $AB$. Choose points $M$ and $N$ on sides $AC$ and $BC$ respectively such that $DM$ is parallel to $BC$ and $DN$ is parallel to $AC$. If $DM=5$ and $DN=4$,then $AC$ and $BC$ are respectively equal to:

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