To find two sides of a triangle when it is circumscribed a circle

Question

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

I tried to solve this problem by my way but i failed to do so i need a short and understandable solution of this problem. Thanks

Answer 1

Finding the numbers

If all I needed was the answer, I'd simply construct this in e.g. Cinderella:

Using a formula

If you need a formula, then look at the Wikipedia article on incircles. In the section “Other incircle properties” you'll find this:

Suppose the tangency points of the incircle divide the sides into lengths of $x$ and $y$, $y$ and $z$, and $z$ and $x$. Then the incircle has the radius^[11] $$r=\sqrt{\frac{xyz}{x+y+z}}$$

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^[11] Chu, Thomas, The Pentagon, Spring 2005, p. 45, problem 584.

So you have $r=4,y=8,z=6$ which gives you

\begin{align*} 4^2&=\frac{x\cdot 8\cdot 6}{x+8+6}\\ 16(x+14)&=48x \\ x+14&=3x \\ 2x&=14 \\ x&=7 \end{align*}

From which you can find $AB=8+x=15$ and $AC=6+x=13$ as in the figure.

Finding that formula

I haven't read or even found the cited reference, but you could also find that formula using a bit of trigonometry. Start with

$$\tan\angle OAB=\tan\frac\alpha2=\frac{OE}{AE}=\frac rx$$

and similar for the other two half angles. Using the double angle formula for the tangent you get

$$\tan\alpha=\frac{2\frac rx}{1-\left(\frac rx\right)^2}=\frac{2rx}{x^2-r^2}$$

Then using the formula for angle sums you get

$$ \tan(\alpha+\beta)=\frac{\frac{2rx}{x^2-r^2}+\frac{2ry}{y^2-r^2}}{1-\frac{2rx}{x^2-r^2}\cdot \frac{2ry}{y^2-r^2}}=\frac{2rx(y^2-r^2)+2ry(x^2-r^2)}{(x^2-r^2)(y^2-r^2)-4r^2xy} %=\frac{2r(x+y)(xy-r^2)}{(xy-rx-ry-r^2)(xy+rx+ry-r^2)} \\ \tan(\alpha+\beta+\gamma)=\frac{\frac{2rx(y^2-r^2)+2ry(x^2-r^2)}{(x^2-r^2)(y^2-r^2)-4r^2xy}+\frac{2rz}{z^2-r^2}}{1-\frac{2rx(y^2-r^2)+2ry(x^2-r^2)}{(x^2-r^2)(y^2-r^2)-4r^2xy}\cdot\frac{2rz}{z^2-r^2}} \\=\frac{(2rx(y^2-r^2)+2ry(x^2-r^2))(z^2-r^2)+2rz((x^2-r^2)(y^2-r^2)-4r^2xy)}{((x^2-r^2)(y^2-r^2)-4r^2xy)(z^2-r^2)-(2rx(y^2-r^2)+2ry(x^2-r^2))2rz} $$

Since $\tan(180°)=0$ you know that the above must be zero, which means that its numerator must be zero. Factorizing that numerator (using a computer algebra system, e.g. Wolfram Alpha), you get

$$ 2\cdot r \cdot (r^{2} - x y - x z - y z) \cdot (r^{2} x + r^{2} y + r^{2} z - x y z) = 0 $$

One of these two parentheses must encode the relation we are looking for. To find out which of the two it is, you could consider the special case of the regular triangle, where $x=y=z$. The first parenthesis would lead to $r=\sqrt3x$ which is larger than the incircle. The second has $r=\frac13\sqrt3x$ which fits much better. That second parenthesis can be reformulated to the form stated in Wikipedia.

Answer 2

Hint:-

$\dfrac{1}{2}OD(CB+BA+AC)=\sqrt{S(S-AC)(S-CB)(S-BA)}$

$$S=\dfrac{CB+BA+AC}{2}$$