Question
In $\triangle\text{ABC},$ AD and BE are altitudes. Prove that: $\frac{\text{ar}(\triangle\text{DEC})}{\text{ar}(\triangle\text{ABC})}=\frac{\text{DC}^2}{\text{AC}^2}$
✓
Answer
Given: $\triangle\text{ABC}$ in which AD and BE are altitudes on sides BC and AC respectively.Since $\angle\text{ADB}=\angle\text{AEB}=90^\circ,$ there must be a circle passing through point D and E having AB as diameter.
We also know that, angle in a semi-circle is a right angle.
Now, join DE.
So ABDE is a cyclic quadrilateral with AB being the diameter of the circle.
$\angle\text{A}+\angle\text{BDE}=180^\circ$ [Opposite angles in a cyclic quadrilateral are supplementary]
$\Rightarrow\angle\text{A}+(\angle\text{BDA}+\angle\text{ADE})=180^\circ$
$\Rightarrow\angle\text{BDA}+\angle\text{ADE}=180^\circ-\angle\text{A}\ ....(1)$
Again
$\angle\text{BDA}+\angle\text{ADC}=180^\circ$ [Linear pair]
$\Rightarrow\angle\text{BDA}+\angle\text{ADE}+\angle\text{EDC}=180^\circ$
$\Rightarrow\angle\text{BDA}+\angle\text{ADE}=180^\circ-\angle\text{EDC}\ \ ...(2)$
Equating (1) and (2), we get
$180^\circ-\angle\text{A}=180^\circ-\angle\text{EDC}$
$\Rightarrow\angle\text{A}=\angle\text{EDC}$
Similarly, $\angle\text{B}=\angle\text{CED}$
Now, in $\triangle\text{ABC}$ and $\triangle\text{DEC}$, we have
$\angle\text{A}=\angle\text{EDC}$
$\angle\text{B}=\angle\text{CED}$
$\angle\text{C}=\angle\text{C}$
$\therefore\triangle\text{ABC}\sim\triangle\text{DEC}$
$\Rightarrow\frac{\text{Area of }\triangle\text{DEC}}{\text{Area of }\triangle\text{ABC}}=\Big(\frac{\text{DC}}{\text{AC}}\Big)^2$
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