D and E are the mid points of two cords AB and AC of a circle with a centre O. The line OD and OE are produced to meet the circle respectively at the point F and G. H is a point on the major arc between F and G of circle. If ∠BAC = 40o , then ∠ FHG=? O केंद्र वाले वृत्त की दो जीवाएं AB और AC के मध्य बिंदु D और E है |रेखाएं OD और OE आगे बढ़ते हुए वृत्त के बिंदु F और G पर मिलती है | वृत्त के वृहद् चाप F और G पर एक बिंदु H है | यदि ∠BAC = 40o , तब ∠ FHG=?

D is the midpoint of chord AB, so OD is perpendicular to AB (property: line from center to midpoint of chord is perpendicular to chord). Similarly, OE is perpendicular to AC. This means ∠BAC = 40° is an inscribed angle subtending arc BC. Central angle ∠BOC = 2 × ∠BAC = 80°. Points F and G are on the circle, and H is on the major arc FG. Using cyclic quadrilateral properties, ∠FHG = (180° - ∠FOG)/2. After calculation: ∠FHG = 180° - ∠F O G/2 = 180° - 40° = 140°, then using chord properties: ∠FHG = 70°.