# Help me verify whether my solution of this geometry problem is correct

Consider the following problem.

# The Problem

Point $$B$$ lies on the segment $$AC$$. A tangent line is constructed from point $$A$$ to the circle of diameter $$BC$$, intersecting it at point $$M$$. $$K$$ is the second point of intersection of line $$AM$$ with the circle of diameter $$AB$$. The extension of segment $$MB$$ intersects the circle of diameter $$AB$$ at point $$D$$.

a) Prove $$AD\parallel MC$$.

b) Find the area of $$DBC$$ if $$AK=5$$ and $$MK=25$$.

# My Solution

a) $$\angle ADB = \angle BMC = 90^\circ$$ as inscribed angles subtended by the diameter are always right. As $$\angle ADB$$ and $$\angle BMC$$ are alternate interior angles of a transversal intersecting two lines $$AD$$ and $$MC$$, and as they are equal, then, by Proposition 27 of Euclid (one of criterion for parallel lines) these lines are parallel, i.e. $$AD\parallel MC$$, QED.

b) Because $$AD\parallel MC$$, $$DAMC$$ is a trapezoid with bases $$AD$$ and $$MC$$.

As per properties of a trapezoid, its diagonals divide its interior region into four triangles: two triangles that include one of this trapezoid's bases are similar, and two other triangles have the same area. Thus, to find the area of triangle $$_\Delta DBC$$, it's sufficient to find the area of a triangle $$_\Delta ABM$$, as their areas are equal.

In $$_\Delta ABM$$, we can immediately find side $$AM=5+25=30$$. We now find the altitude to that side in order to find out the area of this triangle. This altitude is $$BK$$, as $$\angle AKB = 90^\circ$$ as an inscribed angle subtended by the diameter.

Let $$\angle AMB = \alpha$$. This is the angle between tangent $$AM$$ and chord $$MB$$, thus, by the alternate segment theorem, it is equal to the angle in the alternate segment $$BM$$. Thus, $$\angle BCM = \angle AMB = \alpha$$. Then $$\angle CBM=90^\circ - \alpha$$ as triangle $$CMB$$ is a right triangle (established in a) earlier), and sum of acute angles in a right triangle is always $$90^\circ$$.

Now, $$\angle KMB=\angle ABM$$ because point $$K$$ lies on a line $$AM$$ between points $$A$$ and $$M$$. Therefore, $$\angle KBM=\angle CBM=\alpha$$. Because triangle $$MKB$$ is also a right triangle, then $$\angle MBK=90^\circ-\alpha=\angle CBM$$. Thus, by axiom of angle measurement, $$\angle CBK=\angle MBK+\angle CBM=90^\circ-\alpha+90^\circ-\alpha=180^\circ-2\alpha$$. Now, as $$\angle ABK$$ is adjacent to $$\angle CBK$$, it's true that $$\angle ABK+\angle CBK=180^\circ$$. Therefore, $$\angle ABK=2\alpha$$.

As triangle $$AKB$$ is a right triangle as $$BK\perp AK$$, it's true that $$\angle BAK=90^\circ-2\alpha$$.

Let's now express $$\tan\alpha$$ in terms of some sides. By definition of a tangent of an acute angle of a right triangle, it's the ratio of the opposite leg to the adjacent leg. Therefore, from triangle $$MKB$$, $$\tan\alpha=\dfrac{BK}{MK}=\dfrac{BK}{25}$$.

By one of the phase shift identities, $$\tan(90^\circ - 2\alpha)=\cot 2\alpha$$. By the double-angle identity for cotangent, $$\cot 2\alpha=\dfrac{1-\tan^2\alpha}{2\tan\alpha}$$. From the right triangle $$ABK$$ and the definition of tangent, we also have $$\tan(90^\circ - 2\alpha)=\tan\angle BAK=\dfrac{BK}{AK}=\dfrac{BK}{5}$$. Therefore, it's true that $$\cot 2\alpha=\dfrac{BK}{5}$$.

Let $$\tan\alpha = x$$, and $$BK = y$$. We now solve the system of equations from what we have got earlier:

$$\begin{cases} x=\dfrac{y}{25}, \\ \dfrac{1-x^2}{2x} = \dfrac{y}{5}; \end{cases} \Rightarrow \begin{cases} y=25x, \\ y=5\cdot\dfrac{1-x^2}{2x}; \end{cases} \Rightarrow 25x = 5\cdot\dfrac{1-x^2}{2x}.$$

We now solve the equation for $$x$$:

\begin{align*} 25x &= 5\cdot\dfrac{1-x^2}{2x} \\ 5x &= \dfrac{1-x^2}{2x} \\ 10x^2 &= 1-x^2 \\ 11x^2 &= 1 \\ x^2 &= \dfrac{1}{11} \end{align*}

That means $$\tan^2\alpha=\dfrac{1}{11}$$. We know $$\alpha$$ is an acute angle, so $$\tan\alpha > 0$$. Then, we will take the positive root: $$\tan\alpha=\dfrac{1}{\sqrt{11}}$$.

Now we can find $$BK$$.

\begin{align*} \tan\alpha &= \dfrac{BK}{25} \\ \dfrac{BK}{25} &= \dfrac{1}{\sqrt{11}} \\ BK &= \dfrac{25}{\sqrt{11}} \end{align*}

We can now find the area of a triangle $$_\Delta ABM$$:

$$S_{_\Delta ABM} = \dfrac{1}{2}\cdot AM\cdot BK = \dfrac{1}{2}\cdot 30\cdot \dfrac{25}{\sqrt{11}} = \dfrac{375}{\sqrt{11}}$$

Therefore, as $$S_{_\Delta ABM}=S_{_\Delta DBC}$$, we conclude that $$S_{_\Delta DBC}=\dfrac{375}{\sqrt{11}}$$.

# My Question

Is my solution correct, or does it have a flaw? The other person got an answer $$\dfrac{75\sqrt{11}}{2}$$. Which answer is correct, and why?

• Your solution is correct. Commented Jun 5, 2023 at 15:17
• @Rusuano, you figure does not match the statement, please change it. Commented Jun 6, 2023 at 5:00
• @sirous I'll redo the drawing in GeoGebra in order to meet the quality guidelines. Commented Jun 6, 2023 at 7:12

Let $$O$$ be the midpoint of $$BC$$. Angle chasing yields $$MO\parallel KB$$ (I leave you this part as an exercise). So, if $$\overline{BC} = 2x$$, $$\overline{AB} = \frac15 x$$, and $$\overline{KB} = \frac16 x$$.
Pythagorean Theorem on $$\triangle AKB$$ gives $$\frac1{25}x^2 = 25+\frac1{36}x^2,$$ that is $$x=\frac{150}{\sqrt{11}}.$$
Again Pythagoras on $$\triangle KBM$$ and $$\triangle BCM$$ yields $$\overline{MB} =\frac{50\sqrt{33}}{11},$$ and $$\overline{MC} = 50\sqrt 3$$ Similarity $$\triangle ADM \sim \triangle MBC$$ gives $$\overline{DB} = \frac{5\sqrt{33}}{11}$$ and $$\overline{AD} = 5\sqrt 3.$$ If $$DH$$ is the line segment perpendicular to $$AB$$ we get $$\overline{DH} = \frac{\overline{AD}\cdot\overline{DB}}{\overline{AB}}.$$ Thus $$\mathcal A_{DBC} = \frac{\overline{BC}\cdot\overline{AD}\cdot\overline{DB}}{2\overline{AB}}=\frac{375}{\sqrt{11}}.$$
• The 'angle chasing' you've been mentioning is likely this: let $\angle KMB=\alpha$, then by the alternate segment theorem, $\angle KMB=\angle MCB=\alpha$. $MO$ will be the median of hypotenuse $BC$, thus $OM=OB=OC$, which yields $\angle OCM = \angle OMC = \alpha$. Therefore, $\angle OMB=90^\circ - \alpha$, and finally, $\angle OMK=90^\circ-\alpha+\alpha=90^\circ$, which leads to $OM\perp AM$, and since $BK\perp AM$ too, then $OM\parallel BK$, QED. Commented Jun 6, 2023 at 7:24