The point of concurrency of three altitudes of a triangle is called as:

In the given Fig., $DE\parallel BC, AD=2 \mathrm{cm}, DB=3 \mathrm{cm} \mathrm{and} AE=1.6 \mathrm{cm}$. Then $EC$ is equal to:

In the given Fig., $DE\parallel BC, AD=2.4 \mathrm{cm}, AE=3.2 \mathrm{cm} \mathrm{and} CE=4.8 \mathrm{cm}$. Then the value of $BD$ is equal to:

In the given figure, $DE$ is parallel to $BC$, $\frac{AD}{DB}=\frac{5}{3}$. If $AE=7.8 \mathrm{cm},$ the value of $EC$ is equal to: 

The sides of two similar triangles are in the ratio $4:9$. Areas of these triangles are in the ratio:

What value(s) of $x$ will make $DE\parallel AB$ in the following figure?

In the fig., if $\angle ADE=\angle B,$ such that $△ADE~△ABC$. If $AD=6.8 \mathrm{cm}, AE=8.6 \mathrm{cm}, BE=2.4 \mathrm{cm} \mathrm{and} BC=5.5 \mathrm{cm}$, find $DE$.