Someone check if my solution is correct or not. First we prove this for quadrilateral. Suppose $ABCD$ is a quadrilateral with interior diagonal $AC$. Assume for contrary, $ABCD$ is concave. But $\angle ABC + \angle ADC>180^{\circ}\implies \angle BAD,\angle BCD<180^{\circ}$ which is a contradiction. So, $ABCD$ is convex. We can also check that, The circumcircles of $\Delta ABC$ and $\triangle ACD$ contains the entire quadrilateral as $\angle ABC + \angle ADC>180^{\circ}$. Now we will prove for polygons with more than $4$ sides. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -0.5495116453794103, xmax = 15.408414725770077, ymin = -3.149496619083398, ymax = 5.926386175807665; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((1.02,3.5)--(6.36,0.56)--(4.4,4.42)--(3.02,3.16)--cycle, linewidth(0.8) + rvwvcq); draw((8.1,0.98)--(12.62,-1.6)--(11.1,3.5)--(9.96,1.62)--cycle, linewidth(0.8) + rvwvcq); draw((12.86,3.5)--(12.62,-1.6)--(14.4,0)--(13.02,1.4)--cycle, linewidth(0.8) + rvwvcq); /* draw figures */ draw((1.02,3.5)--(0.48,0.92), linewidth(0.8) + wrwrwr); draw((0.48,0.92)--(2.02,-1.02), linewidth(0.8) + wrwrwr); draw((2.02,-1.02)--(5.08,-1.28), linewidth(0.8) + wrwrwr); draw((5.08,-1.28)--(6.36,0.56), linewidth(0.8) + wrwrwr); draw((6.36,0.56)--(6.22,2.98), linewidth(0.8) + wrwrwr); draw((6.22,2.98)--(4.4,4.42), linewidth(0.8) + wrwrwr); draw((4.4,4.42)--(3.02,3.16), linewidth(0.8) + wrwrwr); draw((3.02,3.16)--(1.02,3.5), linewidth(0.8) + wrwrwr); draw((0.48,0.92)--(5.08,-1.28), linewidth(0.8) + wrwrwr); draw((1.02,3.5)--(6.36,0.56), linewidth(0.8) + wrwrwr); draw((5.08,-1.28)--(1.02,3.5), linewidth(0.8) + wrwrwr); draw((4.4,4.42)--(6.36,0.56), linewidth(0.8) + wrwrwr); draw((1.02,3.5)--(6.36,0.56), linewidth(0.8) + rvwvcq); draw((6.36,0.56)--(4.4,4.42), linewidth(0.8) + rvwvcq); draw((4.4,4.42)--(3.02,3.16), linewidth(0.8) + rvwvcq); draw((3.02,3.16)--(1.02,3.5), linewidth(0.8) + rvwvcq); draw((8.1,0.98)--(9.96,1.62), linewidth(0.8) + wrwrwr); draw((9.96,1.62)--(11.1,3.5), linewidth(0.8) + wrwrwr); draw((11.1,3.5)--(12.86,3.5), linewidth(0.8) + wrwrwr); draw((12.86,3.5)--(13.02,1.4), linewidth(0.8) + wrwrwr); draw((13.02,1.4)--(14.4,0), linewidth(0.8) + wrwrwr); draw((14.4,0)--(12.62,-1.6), linewidth(0.8) + wrwrwr); draw((12.62,-1.6)--(9.78,-1.22), linewidth(0.8) + wrwrwr); draw((9.78,-1.22)--(8.1,0.98), linewidth(0.8) + wrwrwr); draw((3.02,3.16)--(6.36,0.56), linewidth(0.8) + wrwrwr); draw((8.1,0.98)--(12.62,-1.6), linewidth(0.8) + rvwvcq); draw((12.62,-1.6)--(11.1,3.5), linewidth(0.8) + rvwvcq); draw((11.1,3.5)--(9.96,1.62), linewidth(0.8) + rvwvcq); draw((9.96,1.62)--(8.1,0.98), linewidth(0.8) + rvwvcq); draw((12.86,3.5)--(12.62,-1.6), linewidth(0.8) + rvwvcq); draw((12.62,-1.6)--(14.4,0), linewidth(0.8) + rvwvcq); draw((14.4,0)--(13.02,1.4), linewidth(0.8) + rvwvcq); draw((13.02,1.4)--(12.86,3.5), linewidth(0.8) + rvwvcq); draw((9.96,1.62)--(12.62,-1.6), linewidth(0.8) + wrwrwr); draw((12.62,-1.6)--(13.02,1.4), linewidth(0.8) + wrwrwr); /* dots and labels */ dot((1.02,3.5),dotstyle); label("$A$", (1.0733283245679952,3.6574154770848994), NE * labelscalefactor); dot((0.48,0.92),dotstyle); label("$B$", (0.20180315552216635,1.0879188580015025), NE * labelscalefactor); dot((2.02,-1.02),dotstyle); label("$C$", (1.659353869271225,-1.1209466566491373), NE * labelscalefactor); dot((5.08,-1.28),dotstyle); label("$D$", (5.14545454545454,-1.1359729526671687), NE * labelscalefactor); dot((6.36,0.56),dotstyle); label("$E$", (6.422689706987221,0.7122614575507134), NE * labelscalefactor); dot((6.22,2.98),dotstyle); label("$F$", (6.287453042824937,3.1314951164537947), NE * labelscalefactor); dot((4.4,4.42),dotstyle); label("$G$", (4.45424492862509,4.574019534184825), NE * labelscalefactor); dot((3.02,3.16),dotstyle); label("$H$", (2.8314049586776844,3.4169947407963948), NE * labelscalefactor); dot((9.96,1.62),dotstyle); label("$I$", (9.788580015026284,1.7791284748309544), NE * labelscalefactor); dot((8.1,0.98),dotstyle); label("$J$", (8.165740045078879,1.132997746055597), NE * labelscalefactor); dot((9.78,-1.22),dotstyle); label("$K$", (9.833658903080378,-1.0758677685950424), NE * labelscalefactor); dot((12.62,-1.6),dotstyle); label("$L$", (12.673628850488338,-1.4515251690458315), NE * labelscalefactor); dot((14.4,0),dotstyle); label("$M$", (14.46175807663409,0.1562885048835456), NE * labelscalefactor); dot((13.02,1.4),dotstyle); label("$N$", (13.07933884297519,1.5537340345604809), NE * labelscalefactor); dot((12.86,3.5),dotstyle); label("$O$", (12.914049586776843,3.6574154770848994), NE * labelscalefactor); dot((11.1,3.5),dotstyle); label("$P$", (11.155972952667154,3.6574154770848994), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Assume for contrary, at least one of internal angle of this polygon is greater than $180^{\circ}$ ($H$ on the first diagram). We know that this angle is a part of two triangles that are used in decomposition. Joining them gives us a quadrilateral. By our proof, we know that This quadrilateral is convex which is a contradiction (Like in diagram $1$, $AEGH$ is convex and $IJKP$, $MNOL$ are convex in the second diagram).

Part (b) actually implies part (a) because no circle passing through a concave vertex can contain the whole polygon. Part (b) is easy by induction.