The bisector of an isosceles triangle divides the height drawn to the base

The bisector of an isosceles triangle divides the height drawn to the base, in a ratio of 5/3, find the perimeter of the triangle if this height is 24 cm.

Given:

AB = AC

BD – height

AE – bisector

BO: OD = 5: 3

BD = 24 cm

Find: p?

Solution: Let x be the aspect ratio, then BO = 5x and OD = 3x. Let’s compose and solve the equation:

Find the height BD
5x + 3x = 24;

8x = 24;

x = 3.

From here we get: BO = 15 cm, OD = 9 cm.

At what point do the bisectors intersect
It is known from the planimetry course that:

the height in an isosceles triangle is the bisector and median;
the intersection point of the bisectors is the center of a circle inscribed in a triangle;
OD = 9 cm – we have established this fact above.
OE is the perpendicular to the BC side. OE = 9 cm.

Triangles BDC and OEB are similar – they are two right-angled triangles that share a common angle DBC. From the similarity we have:

In a right triangle BEO

From the above ratio we have that DC = 18 cm.Then AC = 36 cm.

From a right-angled triangle ABD.

BC = AC = 30 cm.

Then the perimeter of triangle ABC = 36 + 30 + 30 = 96 cm.

The same problem can be solved using the tangent. However, this is not entirely correct – at school this topic is discussed later.

Answer: the perimeter of the triangle is 96 cm.