Question

Kindly answer the following questions with diagrams also. Any spam answers will be reported.

Answer 1

Answer:

proved

Step-by-step explanation:

1)Given:-

AB = AC

Also , BD and CE are two medians

Hence ,  

E is the midpoint of AB and

D is the midpoint of CE

Hence ,

1/2 AB = 1/2AC

BE = CD  

In Δ BEC and ΔCDB ,

BE = CD [ Given ]

∠EBC = ∠DCB [ Angles opposite to equal sides AB and AC ]

BC = CB [ Common ]

Hence ,

Δ BEC ≅ ΔCDB [ SAS ]

BD = CE (by CPCT)

2)Given :

Point D and E are on side BC of ∆ABC,

such that BD = CE ,and AD = AE .

To prove : ∆ABD congruent to ∆ACE

proof :

i ) In ∆ADE ,

AD = AE ( given )

<ADE = <AED = x

[ Angles opposite to equal sides ]

ii ) <ADB + <ADC= 180°

[ Linear pair ]

<ADB + x = 180°

=> <ADB = 180° - x -----( 1 )

iii ) Similarly ,

<AEC = 180° - x --------( 2 )

iv ) In ∆ABD and ∆ACE ,

BD = EC ( S ) given ,

<ADE = <AEC ( A ) from ( 1 ) and ( 2 )]

AD = AE ( S ) given ,

Therefore ,

∆ABD is congruent to ∆ACE.

[ ASA congruence Rule ]