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Question Number 3823 by Rasheed Soomro last updated on 21/Dec/15

$$Consideratriangle\mathrm{ABC}.Let\mathrm{D}and\mathrm{E}$$$$aretwopointson\mathrm{AB}and\mathrm{AC}respectively$$$$suchthat\mathrm{DE}\parallel \mathrm{BC}.Nowtherearetwo$$$$partsof△\mathrm{ABC}:△\mathrm{ADE}andtrapizoid$$$$\mathrm{DBCE}.Ifthesetworegionshavesamearea$$$$Whatwillbetheratiooftwodistances:$$$$\left(i\right)distanceof\mathrm{DE}frompoint\mathrm{A}and$$$$\left(ii\right)distancebetween\mathrm{BC}and\mathrm{DE}?$$

Commented by prakash jain last updated on 21/Dec/15

$$\mathrm{AX}-\mathrm{Altitute}\mathrm{from}\mathrm{A}\mathrm{to}\mathrm{BC},\mathrm{intersects}$$$$\mathrm{DE}\mathrm{at}\mathrm{Y}.$$$$△\mathrm{ADY}\mathrm{and}△\mathrm{ABX}\mathrm{are}\mathrm{similar}$$$$\frac{\mathrm{DY}}{\mathrm{BX}}=\frac{\mathrm{AY}}{\mathrm{AX}}=k$$$$△\mathrm{AYF}\mathrm{and}△\mathrm{AXC}\mathrm{are}\mathrm{similar}$$$$\frac{\mathrm{AY}}{\mathrm{AX}}=\frac{\mathrm{YE}}{\mathrm{XC}}=k$$$$\mathrm{YE}=k\mathrm{XC}$$$$\mathrm{DY}=k\mathrm{BX}$$$$\mathrm{DY}+\mathrm{YE}=k(\mathrm{BX}+\mathrm{XC})\Rightarrow \mathrm{DE}=k\mathrm{BC}$$$$\mathrm{AY}=k\mathrm{AX}$$$$\mathrm{XY}=\mathrm{AX}-\mathrm{AY}=(1-k)\mathrm{AX}$$$$\mathrm{area}\mathrm{of}△\mathrm{ADE}=\frac{1}{2}\mathrm{AY}\times \mathrm{DE}=\frac{1}{2}k\mathrm{AX}\times k\mathrm{BC}$$$$\mathrm{area}\mathrm{of}\mathrm{trpezium}\mathrm{DEBC}=\mathrm{BC}\times \mathrm{XY}=\mathrm{BC}\times (1-k)\mathrm{AX}$$$$\frac{1}{2}{k}^2=1-k$$$$k^2-2k-2=0$$$$k=\frac{2\pm \sqrt{4+8}}{2}$$$$k>0\Rightarrow k=1+\sqrt{3}$$

Commented by Rasheed Soomro last updated on 21/Dec/15

$$\mathcal{A}\mathcal{N}ovel\mathcal{M}ethod!$$

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