1. ## Circle Theorems

Hello there,

I'm having problems visualising this question.
I have no idea how to prove it, because I don't know what it would look like in the first place... Could somebody please help?

"The angles of a triangle are 50, 60 and 70 degrees, and a circle touches the sides at A, B, C. Calculate the angles of triangle ABC".

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I have now gotten stuck on another question:

"Line ATB touches a circle at T and TC is a diameter. AC and BC cut the circle at D and E respectively. Prove that the quadrilateral ADEB is cyclic"

The problem is that whenever I draw this, points A and B are outside of the circle and thus it cannot be classified as cyclic. Unless I am wrong...

2. Hello, beefsupreme!

The angles of a triangle are 50°, 60° and 70°,
and a circle touches the sides at $\displaystyle A, B, C.$
Calculate the angles of triangle $\displaystyle ABC$.
Code:
                        P
o
/ \
/50°\
/     \
/       \
/         \
/   * * *   \
/*           *\
*              *
A o- - - - - - - - -o B
/ \               / \
/*  \             /  *\
/ *   \     o     /   * \
/  *    \    O    /    *  \
/         \       /         \
/     *     \     /     *     \
/       *     \   /     *       \
/ 60°      *    \ /    *      70° \
Q o - - - - - - - * o * - - - - - - - o R
C

We have $\displaystyle \Delta PQR$ with $\displaystyle \angle P = 50^o,\,\angle Q = 60^o,\:\angle R = 70^o$

Circle $\displaystyle O$ is inscribed in the triangle
. . and is tangent to $\displaystyle PQ, PR, QR$ at $\displaystyle A,B,C,$ respectively.

Draw segments $\displaystyle OP, OQ, OR$ and radii $\displaystyle OA, OB, OC.$

$\displaystyle OA,OB,OC$ are penpendicular to the sides of the triangle.

Since O is the center of the inscribed circle,
. . $\displaystyle OP, OQ, OR$ bisect $\displaystyle \angle P, \angle Q, \angle R$, resp.

We have: .$\displaystyle \angle APO = 25^o \quad\Rightarrow\quad \angle POA = 65^o \quad\Rightarrow\quad \angle AOB = 130^o$

Since $\displaystyle \Delta AOB$ is isosceles: .$\displaystyle \angle OAB = \angle OBA = 25^o$

Similarly, we find that: .$\displaystyle \angle OAC = \angle OCA = 30^o,\:\angle OBC = \angle OCB = 35^o$

Therefore: .$\displaystyle \angle A = 55^o,\:\angle B = 60^o,\:\angle C = 65^o$