Given that line AD is an altitude of triangle ABC.
Equation to find the altitude AD is:
[tex](AD)^2=(BD)(DC)[/tex]
Where
BD = 3x
DC = 5x - 14
Let's solve for x:
m∠ADC = 5x + 20
m∠DAC = 2x + 2
We know that m∠ADC is 90 degrees
Thus,
5x + 20 = 90
5x = 90 - 20
5x = 70
[tex]\begin{gathered} \frac{5x}{5}=\frac{70}{5} \\ \\ x=14 \end{gathered}[/tex]
Since x = 14, we have:
BD = 3x = 3(14) = 42
DC = 5x - 14 = 5(14) - 14 = 70 - 14 = 56
Therefore
[tex]\begin{gathered} (AD)^2=^{}(BD)(CD) \\ \\ (AD)^2=(42)(56) \\ \\ (AD)^2=2352 \end{gathered}[/tex]
Take the square root of both sides:
[tex]\begin{gathered} \sqrt[]{(AD)^2}=\sqrt[]{2352} \\ \\ AD\text{ = }48.5 \end{gathered}[/tex]
AD = 48.5
ANSWER:
Equation setup = (AD)² = (BD)(DC)
