Question

9. In the figure, A, B, C, and D, E, F are two sets of collinear points. Prove that AD II CF.
10. In the figure, equal chords AB and CD of a circle with centre 0, cut at right angles at E.
If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square.
LIHLZ=180
с
907 N
M
E
B​

Answer 1

Step-by-step explanation:

Step-by-step explanation:Join OE.

Step-by-step explanation:Join OE.

Step-by-step explanation:Join OE.In ΔOME and ΔONE,

Step-by-step explanation:Join OE.In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] 

Step-by-step explanation:Join OE.In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° 

Step-by-step explanation:Join OE.In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] 

Step-by-step explanation:Join OE.In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME  ≅ ∠ONE [by SAS congruency] 

Step-by-step explanation:Join OE.In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME  ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] 

Step-by-step explanation:Join OE.In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME  ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN,

Step-by-step explanation:Join OE.In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME  ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ∠MON = 360° - (∠OME + ∠MEN + ∠ONE)

Step-by-step explanation:Join OE.In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME  ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ∠MON = 360° - (∠OME + ∠MEN + ∠ONE) = 360° - (90° + 90° + 90°) = 90° [∠MEN = 90°, given]

Step-by-step explanation:Join OE.In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME  ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ∠MON = 360° - (∠OME + ∠MEN + ∠ONE) = 360° - (90° + 90° + 90°) = 90° [∠MEN = 90°, given] Thus, in quadrilateral OMEN, 

Step-by-step explanation:Join OE.In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME  ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ∠MON = 360° - (∠OME + ∠MEN + ∠ONE) = 360° - (90° + 90° + 90°) = 90° [∠MEN = 90°, given] Thus, in quadrilateral OMEN, OM =ON , ME = NE 

Step-by-step explanation:Join OE.In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME  ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ∠MON = 360° - (∠OME + ∠MEN + ∠ONE) = 360° - (90° + 90° + 90°) = 90° [∠MEN = 90°, given] Thus, in quadrilateral OMEN, OM =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90°

Step-by-step explanation:Join OE.In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME  ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ∠MON = 360° - (∠OME + ∠MEN + ∠ONE) = 360° - (90° + 90° + 90°) = 90° [∠MEN = 90°, given] Thus, in quadrilateral OMEN, OM =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved. 

Answer 2

Answer:

ab aap kon ho yarr abhi ko mera no . hi kyu chahiye