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Question

Internal bisector of $$\angle \mathrm{A}$$ of triangle ABC meets side BC at D. A line drawn through $$\mathrm{D}$$ perpendicular to AD intersects the side AC at $$\mathrm{E}$$ and the side AB at F. If $$\mathrm{a},\ \mathrm{b},\ \mathrm{c}$$ represent sides of $$\Delta \mathrm{A}\mathrm{B}\mathrm{C}$$ then 


A
AE is HM of b and c
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B
AD =2bcb+ccosA2
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C
EF=4bcb+csinA2
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D
the triangle AEF is isosceles
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Solution

The correct options are
A AE is HM of b and c
B EF$$=\displaystyle \frac{4\mathrm{b}\mathrm{c}}{\mathrm{b}+\mathrm{c}}\sin\frac{\mathrm{A}}{2}$$
C the triangle AEF is isosceles
D AD $$=\displaystyle \frac{2\mathrm{b}\mathrm{c}}{\mathrm{b}+\mathrm{c}}\cos\frac{\mathrm{A}}{2}$$
We have $$\Delta \mathrm{A}\mathrm{B}\mathrm{C}=\Delta \mathrm{A}\mathrm{B}\mathrm{D}+\Delta \mathrm{A}\mathrm{C}\mathrm{D}$$ 

$$\Rightarrow \displaystyle \frac{1}{2}$$ $$bc$$ $$\displaystyle \sin \mathrm{A}=\frac{1}{2}\mathrm{c}\mathrm{A}\mathrm{D}\sin\frac{\mathrm{A}}{2}+\frac{1}{2}\mathrm{b}\times \mathrm{A}\mathrm{D}\sin\frac{\mathrm{A}}{2} \Rightarrow$$ $$AD$$ $$=\displaystyle \frac{2\mathrm{b}\mathrm{c}}{\mathrm{b}+\mathrm{c}}\cos\frac{\mathrm{A}}{2}$$

Again $$AE$$ $$=$$ $$AD$$ $$\displaystyle \sec\frac{\mathrm{A}}{2}$$

$$=\displaystyle \frac{2\mathrm{b}\mathrm{c}}{\mathrm{b}+\mathrm{c}}\Rightarrow$$ $$AE$$ is HM of $$\mathrm{b}$$ and $$\mathrm{c}$$.
$$EF $$ $$=$$$$ ED + DF $$$$=2\mathrm{D}\mathrm{E}=2\times AD $$

$$\displaystyle \tan\frac{\mathrm{A}}{2}=\frac{2\times 2\mathrm{b}\mathrm{c}}{\mathrm{b}+\mathrm{c}}\times\cos\frac{\mathrm{A}}{2}\times\tan\frac{\mathrm{A}}{2}$$

$$=\dfrac{4\mathrm{b}\mathrm{c}}{\mathrm{b}+\mathrm{c}}\sin\dfrac{\mathrm{A}}{2}
$$ As $$AD$$ $$\perp$$ $$EF$$ and $$DE$$$$ =$$$$ DF$$ and $$AD$$ 
is bisector $$\Rightarrow$$ $$AEF$$ is isosceles. 
Hence $$\mathrm{A},\ \mathrm{B},\ \mathrm{C}$$ and $$\mathrm{D}$$ are correct answers. 

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