Question

pls solve this assignment its urgent pls pls pls pls pls pls pls

Answer 1

Hello Mate!

19. [ Error correction - It will be other 'non - parallel' side instead of diagonal ]

Given : ABCD is trapezium where XZ bisects BC and parallel to CD.

To prove : XZ bisect AD

To construct : Join BD

Proof : When XY || CD and X is mid point on BC therefore,

Y is also mid point on BD.

Now, if XZ || CD and AB || CD so XY || AB

Therefore, if YZ is || AB and Y is mid point on BD then Z is mid point on AD.

Hence Proved

20. Given : In ∆ABC, AD is median on BC and BE is median on AD.

To construct : Draw DE || EF

To prove : AF = ⅓ AC

Proof : In ∆ADK, since E is mid point on AD and EF || DK, so F is mid point on AK

AF = FK

In ∆ BCF, since Di is mid point and BF || DK so K is mid point on CF

FK = CK

Therefore, AF = FK = CK

AF + FK + CK = AC

AF + AF + AF = AC

3AF = AC

AF = ⅓ AC

Hence proved

21. Given : In right ∆ABC, M is mid point at AB

To prove : (i) D is also mid point on AC
(ii) CM = AM = ½ AB

Proof : (i) When M is mid point and DM || BC so D is also mid point on AC

(ii) In ∆ADM and ∆CDM,

DM = DM ( Common )

< ADM = < CDM ( 90° each )

AD = CD

Hence both ∆s are ~ by SAS congruency

So AM = CM

AM = ½ AB

AM = CM = ½ AB

Hence proved

22. Given : ABCS is ||gm whose angle bisectors meet to form EFGH.

To prove : EFGH is rectangle

Proof : < BAD + < ADC = 180°

½ <BAD + ½ <ADC = 90°

< DAH + < ADH = 90°

In ∆ ADH

< DAH + < ADH + < AHD = 180°

90° + < AHD = 180°

< AHD = 90°

< AHD = < GHE = 90° ( Vertically opp. angles ]

Similarly, < HGF = < GFE = < HEF = 90°

Hence, all angles are 90° so EFGH is rectangle.

Hence proved

Have great future ahead!