The height AD of an acute-angled triangle ABC divides its base into two segments such that BD

| August 16, 2021 | education

The height AD of an acute-angled triangle ABC divides its base into two segments such that BD = 8 and DC = 9. Find the length height AD if it is known that the height of the triangle BK divides the height AD in half.

Consider a triangle ABC with heights AD and BK lowered to the sides BC and AC.

Let us denote the point of intersection of the heights by O.

By the condition of the problem, it is known that:

BD = 8, DC = 9 and AO = DO.

Consider a triangle BOD. It is obviously rectangular.

Therefore, we have:

DO = BD * tg (OBD) = 8 * tg (OBD).

Note that the angles ОВD = DАС, since triangles АСD, ВКС – rectangular and they have a common angle АСВ.

Therefore, we have:

DC = AD * tg (DAС),

9 = 2 * DO * tan (OBD) = 2 * 8 * tan ^ 2 (OBD),

tg (OBD) = 3/4. From here we get:

9 = AD * 3/4,

AD = 12.

Answer: AD = 12.



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