#### Answer

As we know
Area $\triangle$ ABC = Area $\triangle$ ADC + Area $\triangle$ BCD
Calculate the area of the triangles and used the values in the above equation.
Now solve the equation you get the desired result.

#### Work Step by Step

Given right $\triangle$ ABC
We need to prove h = $\frac{ab}{c}$
Let us take a point which intersect Ab and Let AD = x
AB= AD + BD
c = x + BD
BD = c - x
Area $\triangle$ ABC = Area $\triangle$ ADC + Area $\triangle$ BCD
The area of right triangle with legs of length a and b given by A = $\frac{1}{2}$ ab
area of triangle ABC = $\frac{1}{2}$ ab
Area of triangle ADC = $\frac{1}{2}$ xh
Area of triangle BCD = $\frac{1}{2}$(c-x)h
Substitute the values of Area in
Area $\triangle$ ABC = Area $\triangle$ ADC + Area $\triangle$ BCD
$\frac{1}{2}$ ab = $\frac{1}{2}$ xh + $\frac{1}{2}$(c-x)h
ab = xh + (c-x)h
ab = ch
h = $\frac{ab}{c}$
Hence Proved