Hard Circle Geometry proofs

Question

I need help with this circle Geometry question, I got a and b but need help with c.

Answer 1

(i) is easy and acts as a hint to the parts that follow.

(ii) $\angle GBF = \angle GCA$ (angles in the same segment)

$ \angle GCA = \angle HDF$ (for the same reason and is a result from (i))

$ \angle HDF = \angle HBF$ (for the same reason as above).

(iii) Both chords subtends equal angles (ie. $\angle BCG = \angle BAK$)

Added: If two chords of the same circle are respectively subtended by the same sized angle(s) on the circumference, these two chords must be equal in length.

Answer 2

i. As $\angle BFH=\angle BDH=\frac{\pi}{2}$, $BFHD$ is cyclic with $(BH)$ being the diameter of the circumscribed circle. Similarly, as $\angle CFA=\angle CDA=\frac{\pi}{2}$, $AFDC$ is cyclic with $(AC)$ being the diameter of the circumscribed circle.

ii. $\angle GBF\equiv\underbrace{\angle GBA=\angle GCA}_{\bigcirc ABC}\equiv \underbrace{\angle FCA=\angle FDA}_{\bigcirc AFDC}\equiv \underbrace{\angle FDH=\angle FBH}_{\bigcirc BFHD}$.

iii. $(BF)$ is both angle bisector and altitude of $\triangle GBH \Rightarrow$ $\triangle GBH$ - isosceles $\Rightarrow BG= BH$. Similarly $\triangle KBH$ - isosceles $\Rightarrow BH=BK$.