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Question Number 215769 by momoga last updated on 17/Jan/25

In △ABC, it is given that AC⊥CB, CD⊥AB, and CD = 12, AC = BC + 5. Please solve for the value of BC using a purely geometric method.

Commented by momoga last updated on 17/Jan/25

Answered by A5T last updated on 17/Jan/25

$$\mathrm{AD}=\sqrt{{(\mathrm{BC}+5)}^2-{12}^2}$$$$\mathrm{BD}=\sqrt{{\mathrm{BC}}^2-{12}^2}$$$$\mathrm{AD}+\mathrm{BD}=\sqrt{{(\mathrm{BC}+5)}^2+{\mathrm{BC}}^2}$$$$\Rightarrow {(\mathrm{AD}+\mathrm{BD})}^2={\mathrm{AD}}^2+{\mathrm{BD}}^2+2\mathrm{AD}\times \mathrm{BD}$$$$\Rightarrow 144=\sqrt{({\mathrm{BC}}^2-{12}^2)({\mathrm{BC}}^2+10\mathrm{BC}-119)}$$$$\Rightarrow {144}^2=(\mathrm{BC}-12)(\mathrm{BC}+12)(\mathrm{BC}+17)(\mathrm{BC}-7)$$$$\Rightarrow \mathrm{BC}=15$$

Commented by momoga last updated on 18/Jan/25

Oh, so that’s how it is

Commented by momoga last updated on 17/Jan/25

I believe this is an algebraic approach, involving solving a quartic equation. Is there a geometric method, perhaps?

Commented by AntonCWX last updated on 18/Jan/25

$$No.$$$$Whenwearesolvingthiskindofproblem,$$$$Weapplyconceptfromgeometryandtransformitintoequations.$$

Commented by AntonCWX last updated on 18/Jan/25

$$Theonlypuregeometricwayistodrawthediagraminscaleandmeasurethelength.$$

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