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S seema garhwal
In,               ( given )               (common )       ( By AA rule)     ( corresponding sides of similar triangles )

S seema garhwal
Let AB = 1.8 m BC is horizontal distance between fly to the tip of the rod. Then, the length of the string is AC. In  ABC, using Pythagoras theorem Hence, the length of the string which is out is 3m. If she pulls in the string at the rate of 5cm/s, then the distance travelled by fly in 12 seconds.                           =  Let D be the position of fly after 12 seconds. Hence, AD is the...

S seema garhwal
Produce BA to P, such that AP=AC and join P to C.          (Given ) Using converse of Thales theorem, AD||PC (Corresponding angles)            (Alternate angles) By construction, AP=AC From equation 1,2,3, we get Thus, AD bisects angle BAC.

S seema garhwal
In           (Common)   (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle) So,         ( By AA rule) 24440             (Corresponding sides of similar triangles are proportional)

S seema garhwal
In           (Common)   (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle) So,         ( By AA rule)

S seema garhwal
Join BC In          ( vertically opposite angle)       (Angles in the same segment)          (By AA)              (Corresponding sides of similar triangles are proportional)

S seema garhwal
Join BC In          ( vertically opposite angle)       (Angles in the same segment)          (By AA)

S seema garhwal
In parallelogram ABCD, AF and DE are altitudes drawn on DC and produced BA. In DEA, by Pythagoras theorem In DEB, by Pythagoras theorem ....................................2 In ADF, by Pythagoras theorem In AFC, by Pythagoras theorem Since ABCD is a parallelogram. SO, AB=CD  and BC=AD In        (AE||DF) AD=AD  (common)         (ASA rule) Adding 2 and, we get    (From 4 and...

S seema garhwal
In ABM, by Pythagoras theorem In AMC, by Pythagoras theorem ..................................2 Adding equation 1 and 2,

S seema garhwal
In ABM, by Pythagoras theorem                        (BC=2 BD)

S seema garhwal
Given: AD is a median of a triangle ABC and  AM  BC. In AMD, by Pythagoras theorem In AMC, by Pythagoras theorem             (From 1)            (BC=2 DC)

S seema garhwal
In ADB, by Pythagoras theorem In ACD, by Pythagoras theorem          (From 1)

S seema garhwal
In ADB, by Pythagoras theorem In ACD, by Pythagoras theorem             (From 1)

S seema garhwal
In  DBN,  In  DAN,  BD AC,  From equation 1 and 3, we get  From equation 2 and 3, we get  In       (By AA)            (NB=DM) Hence proved

S seema garhwal
Join BD Given :  D is a point on hypotenuse AC of D ABC, such that BD  AC, DM  BC and  DN  AB.Also DN || BC, DM||NB In  CDM,  In  DMB,  From equation 1 and 2, we get  From equation 1 and 3, we get  In       (By AA)            (BM=DN) Hence proved

S seema garhwal
A line RT is drawn parallel to SP which intersect QP produced at T. Given: PS is the bisector of . By construction,    (as PS||TR)    (as PS||TR) From the above equations, we get By construction, PS||TR In QTR, by Thales theorem, Hence proved

S seema garhwal
In   AB =   cm, AC = 12 cm and BC = 6 cm.                                                                                      It satisfies the pythagoras theorem. Hence, ABC is a right angled triangle and right angled at B. Option C is correct.

S seema garhwal
Given: An equilateral triangle ABC. Let AB=BC=CA=a Draw an altitude AE on BC. So,  In AEB, by Pythagoras theorem