ABCD is a square and AC, BD the two diagonals intersect each other at O. If P is a point on AB, such that AO = AP, prove that 3∠POB = ∠AOP.

Hi shravani! student-name Tushar Mehta asked in Math ABCD IS A SQUARE AND AC , BD THE TWO DIGONALS INTERSECT EACH OTHER AT O . IF P IS A POINT ON AB , SUCH THT AO= OP , PROVE THAT 3ANGLE POB = ANGLE AOP 8 Follow 0 student-name Gopal.mohanty... answered this 16194 helpful votes in Math, Class XII-Science Hi Madhuri! Here is the proof to your question. The given information can be represented diagrammatically as: Let ∠POB = x° It is known that diagonals of a square bisect the angles. ∴ ∠OBP = ∠OAP = (90°/2) = 45° Using exterior angle property for ∆OPB, ∠OPA = ∠OBP + ∠POB = 45° + x° In OAP, OA = AP ∴ ∠OPA = ∠AOP ⇒ ∠AOP = 45° + x° … (1) It is known that diagonals of square are perpendicular to each other. ∴ ∠AOP + ∠POB = 90° ⇒ 45° + x° + x° = 90° [Using (1)] ⇒ 2x = 90 – 45 = 45 ⇒ x = 22.5 ∴ ∠POB = 22.5° and ∠AOP = 45° + 22.5° = 67.5° = 3 × 22.5° ⇒ ∠AOP = 3∠POB Hope! You got the proof.