euclidean-geometrygeometryplane-geometry
For reference: In the figure , P and T are points of tangency . Calculate x. (Answer:$90^o$}
Does anyone have any ideas? I couldn't find a way
$PODT$ is cyclic
$\angle PDO \cong \angle DOT\\
\triangle POA :isosceles\\
\triangle OTB: isosceles$
$DO$ is angle bissector $\angle B$
DT = DP
As it has been pointed out in comments, the information provided is not enough. Though according to your diagram, I assume $\small{ AD \perp DF}$ (according to the letters in my drawing)
Let $\small{\angle FOB=2x}$ so $\small{\angle FOE=x}$. And $\small{\angle BCF=x}$
You can clearly see $\small{OADF}$ is a square. Also $\small{\triangle OEF}$ and $\small{\triangle DFG}$ are equivalent. Therefore $\small{\angle FDG=\angle EOF=x}$.
Since $\small{\angle FCD=\angle FDG}$, $\small{\triangle CFD}$ is isosceles.$\small{\implies CF=DF}$
As $\small{OF=DF}$ (sides of the square) $\small{\triangle OCF}$ is equilateral ($\small{\because OF=OC=CF}$, first two as radii of the circle). Immediately you can prove $\small{\triangle OAB}$ is equilateral as well.
$$\implies \angle AOB=60^\circ$$
Using your diagram:
Let's prove that triangle CJI is similar to triangle ACH. HT is perpendicular to JG, but JG is parallel to CI, so that HT and CI are perpendicular. Let those two lines intersect at a point X. Thus, angle IXH = 90 and so is angle IKH, so that IKXH is cyclic and so angle AHC = angle XHK = angle XIK = angle CIJ. Then, angle CJI = angle CJK = 90 - angle JCK = 90 - angle HCM = angle ACH. Thus, they're similar, and we conclude that CJ/AC = CI/AH, ie. that TJ / TA = JG / AH. This implies that triangles TJA and TGH are similar and that JA and GH are parallel - so that angle x is equal to angle AJC which is half of 80 which is 40.
Best Answer
Extend $AP$ and $BT$ to $K$. Then notice that $E$ is an orthocenter of triangle $ABK$ so we only need to prove $D\in KE$.