Prove that the angles between tangents to circles centered on a trapezium are equal

Suppose a trapezium $$ABCD$$. There are circles $$m,n$$ with centres midpoint of leg $$BC=M_{1}$$ and leg $$AD=M_{2}$$, respectively; and diametres $$BC$$ and $$AD$$, respectively. The point $$P$$ is the intersection of $$BC$$ and $$AD$$. We have tangent lines to $$m,n$$ from $$P$$. Prove that the angle $$\alpha$$ between the tangents to $$m$$ is equal to $$\beta$$ between the tangents to $$n$$.

Since the tangents touch the circle only at one point, they are perpendicular to the radii. Therefore, $$\sin\Bigl(\frac{\alpha}{2} \Bigr)=\frac{r}{|M_{1}P|}\implies \alpha=2\sin^{-1}\left(\frac{|CB|}{2|M_{1}P|} \right)~.$$ Likewise for $$\beta$$.

It is then enough to prove that $$\displaystyle\frac{|CB|}{|M_{1}P|}=\frac{|AD|}{|M_{2}P|}$$.

I have had trouble with proving that last part. Your help would be really appreciated!

Say the perpendicular distance from point $$P$$ to the base of trapezium is $$h$$ and it meets the base at $$H$$. Also, perpendicular from the midpoint of the leg $$M_1$$ meets the base at $$H_1$$ and say the perp distance from the midpoint to the base is $$d$$. (Note: given $$AB \parallel CD,$$ perp distance from $$M_2$$ to the base of the trapezium will also be $$d$$ using midpoint theorem)

First note that $$\triangle PHB \sim \triangle M_1H_1B$$

So, $$~ \displaystyle \frac{r_1}{d} = \frac{PM_1 + r_1}{h} \implies PM_1 = \frac{r_1 (h - d)}{d}$$

where $$r_1$$ is radius of the circle on leg $$BC$$.

Now using the right triangle formed by $$P, M_1$$ and the point of tangency to circle $$m$$,

$$\displaystyle \sin \frac{\alpha}{2} = \frac{r_1}{PM_1} = \frac{d}{h-d}$$

Similarly show that $$\displaystyle \sin \frac{\beta}{2} = \frac{d}{h-d}$$

That leads to $$\alpha = \beta$$.

You're very close. Recall that for positive reals (to ensure denominator is non-zero)

$$\frac{a}{b} = \frac{ c}{d} \Leftrightarrow \frac{b-a}{b+a} = \frac{ d-c}{d+c}.$$

So, from OP's work, we have $$\frac{CB/2}{M_1 P } = \frac{AD/2}{M_2 P } \Leftrightarrow \frac{PB } {PC } = \frac { PA} {PD }$$ by applying the above, which is true from the set of similar triangles $$PAB \sim PDC$$ with $$AB\parallel DC$$ .

• A very elegant continuation of my work, thank you very much! Dec 5 '21 at 16:51

Define $$X\in m$$ such that $$PX$$ is tangent to $$m$$, and define $$Y\in n$$ such that $$PY$$ is tangent to $$n$$. Then $$\angle XPM_1=\dfrac{\alpha}{2}$$ and $$\angle YPM_2=\dfrac{\beta}{2}$$, therefore, it is enough to prove that $$\angle XPM_1=\angle YPM_2$$.
Using Thales' Theorem we obtain $$\dfrac{|PB|}{|PC|}=\dfrac{|PA|}{|PD|}$$, let $$k$$ be this common ratio, then $$|PB|=k\cdot |PC|$$ and $$|PA|=k\cdot |PD|$$. Using the Power of a Point Theorem we get$$|PX|^2=|PC|\cdot |PB|=k\cdot |PC|^2$$and$$|PY|^2=|PA|\cdot |PD|=k\cdot |PD|^2.$$Dividing both equations we get$$\left (\frac{|PX|}{|PY|}\right )^2=\left (\frac{|PC|}{|PD|}\right )^2$$i.e. $$\dfrac{|PX|}{|PY|}=\dfrac{|PC|}{|PD|}$$, because the length of a segment is always positive.
Now, $$M_1$$ and $$M_2$$ are the midpoints of $$BC$$ and $$AD$$, and therefore $$AB\parallel M_1M_2\parallel CD$$. Using Thales' Theorem again we get $$\dfrac{|PC|}{|PD|}=\dfrac{|PM_1|}{|PM_2|}$$. Hence, $$\dfrac{|PX|}{|PY|}=\dfrac{|PM_1|}{|PM_2|}$$, i.e. $$\dfrac{|PM_1|}{|PX|}=\dfrac{|PM_2|}{|PY|}$$. Finally, $$\angle PXM_1=90^\circ =\angle PYM_2$$ because $$PX$$ and $$PY$$ are tangents to $$m$$ and $$n$$, putting this together with the fact that $$\dfrac{|PM_1|}{|PX|}=\dfrac{|PM_2|}{|PY|}$$ we obtain that $$\triangle PXM_1$$ and $$\triangle PYM_2$$ are similar, therefore $$\angle XPM_1=\angle YPM_2$$, as wanted.

Alternatively,$$\arccos \left (\frac{\alpha}{2}\right )=\arccos \left (\dfrac{|PM_1|}{|PX|}\right )=\arccos \left (\dfrac{|PM_2|}{|PY|}\right )=\arccos \left (\frac{\beta}{2}\right )$$and again we get $$\dfrac{\alpha}{2}=\dfrac{\beta}{2}$$.

Let use give a proof based on a geometrical transform.

We are in a case where a similarity transformation i.e., a composition of a rotation (angle $$\theta$$) and a homothety (ratio $$r$$) can provide a simple proof. Let us show that there exists a similitude $$S$$ with center $$P$$ such that $$S(A)=B$$ and $$S(D)=C$$.

Knowing that $$P,A,D$$ and $$P,B,C$$ are aligned, it is sufficient to show that

$$\frac{PB}{PA}=\frac{PC}{PD} \ \ \text{which is the ratio } r \tag{1}$$

and (1) is true due to intercept theorem.

Therefore, we can conclude to the equality of angles because this similitude transformation

• sends the circle with diameter $$AD$$ onto the circle with diameter $$BC$$,

• preserves angles, (as any similitude transformation because both the rotation part and the homothety part preserve them).

Remark: the angle $$\theta$$ of the similitude transformation is evidently $$DPC$$.

• A very elegant and pleasing approach! Thank you Dec 5 '21 at 16:51