The angle between common tangent and two chords, which are connecting tangent points and one common point of two circles

Question

enter image description here

So I saw the solving way and it says that the angle BAC equals 90 degrees, and DA=DB=DC, but can't understand why. Can someone help?

Answer 1

In your figure, $\triangle O_1AB$ is isoceles. So, $\angle O_1BA = \angle O_1AB$. But we also have that $\angle O_1 BD = \angle O_1AD = 90^\circ$. Hence $\angle ABD = \angle BAD$ , which makes $\triangle DAB$ isocles.

So $DA = DB$. Similarly $DA = DC$

Here's a nice way of showing that $\angle BAC = 90^\circ$: Note that $\angle BDA = 180^\circ - 2\angle DAB = 180^\circ - 2 (90^\circ - \angle O_1AB) = 2\angle O_1AB = 180^\circ - \angle BO_1A$. Similalry, $\angle ADC = 180^\circ - \angle A O_2C$.

Hence, we get $\angle AO_1B + \angle AO_2C = 180^\circ$ After that, note that $\angle BAC = 180^\circ - \angle BAO_1 - \angle CAO_2 = 180^\circ - \dfrac12\angle AO_1 B -\dfrac12 \angle AO_1C = 180^\circ - 90^\circ = 90^\circ$

I know it's not the best solution, but to be fair I've not angle chased in many, many years.

Answer 2

Hint: Tangents from a point to a circle have equal length.

$DB = DA$ because they are both tangents to $O_1$.
$DC = DA$ because they are both tangents to $O_2$.

Hence $ DA = DB = DC$.

Hint: Well known property of right triangles

By isosceles triangles $\angle DBA = \angle DAB, \angle DAC = \angle DCA$.
Angles in a triangle: $\angle DBA + \angle DAB + \angle DAC + \angle DCA = 180^\circ$.

Hence $ \angle BAC = \angle BAD + \angle DAC = 90 ^ \circ$