In given figure, if CE is perpendicular to AB, CH is perpendicular to GH and CE = CH, show that $AB = GF$.                                                                                                                                                                                                                             [Page No. 104]

In triangles CEA and CHF:

$$CE = CH \quad [\text{Given}]$$

$$CA = CF \quad [\text{Radii of the circle}]$$

$$\angle CEA = \angle CHF = 90^\circ$$

By RHS congruence, $\triangle CEA \cong \triangle CHF$.

So, AE = FH

Since, the perpendicular from the centre to a chord bisects the chord.

Hence, E is the mid-point of AB and H is the mid-point of GF.

So, AB = 2AE and GF = 2FH

Thus, AB = GF