Question

BD BIsects ABC
IF AB=6. BC= 14. AC=14
Find BD

WE KNOW THE ANSWER IS 4.2.
Please show me how to get there

Thank you

Answer 1

Question #38805, Math, Geometry

BD Bisects ABC

IF AB=6. BC= 14. AC=14

Find BD

Solution

If we denote the $\angle ABD$ by $\beta$ then $\angle ABC = 2\beta$. Because $BC = AC$, the triangle $ABC$ is an isosceles one, so we have $\angle ABC = \angle CAB = 2\beta$.

The sine rule states that the sides of a triangle are proportional to the sines of the opposite angles, so from the triangle $ABD$ we obtain the equation (see the Figure):

$$\frac{BD}{\sin \angle CAB} = \frac{AB}{\sin \angle BDA}.$$

The sum of the angles of a triangle is equal $180{}^{\circ}$, so

$$\angle BDA = 180{}^{\circ} - \angle CAB - \angle ABD = 180{}^{\circ} - 3\beta.$$

Thus by substituting the angles expression into the above equation and simplifying we obtain

$$BD = AB \frac{\sin 2\beta}{\sin 3\beta}.$$

Two of the basic compound angle formula is

$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \sin\beta\cos\alpha,$$ $$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\beta\sin\alpha.$$

By substituting $\beta$ for $\alpha$ into the identity (2) first we get

$$\sin 2\beta = 2\sin\beta\cos\beta.$$

By substituting $2\beta$ for $\alpha$ into the formula (2), using the identity (4) and simplifying the result we obtain

$$\sin 3 \beta = \sin \beta (\cos 2 \beta + \cos^ {2} \beta).$$

By putting the equalities (4), (5) into the equation (1) and cancelling the fraction we get

$$B D = A B \frac {2 \cos \beta}{\cos 2 \beta + 2 \cos^ {2} \beta}.$$

Since the triangle $ABC$ is an isosceles, then the altitude $CE$ is the median of the triangle, that is

$$E B = A E = 0. 5 A B = 3.$$

We find from the right triangle $EBC$

$$\cos 2 \beta = \frac {E B}{C B} = \frac {3}{1 4}.$$

Putting $\alpha = \beta$ in the (3) with Pythagorean identity gives

$$\cos 2 \beta = 2 \cos^ {2} \beta - 1.$$

Because the angle $\beta$ is acute we have

$$\cos \beta = \sqrt {\frac {1 + \cos 2 \beta}{2}} = \frac {1}{2} \sqrt {\frac {1 7}{7}}.$$

Finally, by substituting the two obtained values and $AB = 6$ into the equality (6) we find

$$B D = 6 \frac {2 \cdot 0 . 5}{\frac {1 7}{1 4} + \frac {3}{1 4}} \sqrt {\frac {1 7}{7}} = \frac {3}{5} \sqrt {1 1 9} \approx 6. 5 4 5 2.$$

Answer

$$\frac {3}{5} \sqrt {1 1 9}$$