Question

30. D is the mid-point of side BC of ABC and E is the mid-point of AD. BE produced meets
AC at the point M. Prove that BE - 3EM.
hey guys plz help me n kindly b fast!! ​

Answer 1

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GIVEN : In ∆ABC , D is the midpoint of BC, E is the midpoint of AD. BE produced meets AC at X.

To prove :  BE: EX = 3 : 1

PROOF :

In Δ BCX  and  ΔDCY

∠CBX  = ΔCBY   (corresponding angles)

∠CXB  = ΔCYD   (corresponding angles)

ΔBCX∼ΔDCY   (AA similarity)

BC/DC = BX/ DY = CX/CY

[Since, corresponding sides of two similar triangles are proportional]

BX/DY = BC/DC

BX/DY = 2DC/DC  

[As D is the midpoint of BC]

BX/DY = 2/1………...(i)

In ΔAEX and ΔADY,

∠AEX  = ΔADY   (corresponding angles)

∠AXE = ΔAYD   (corresponding angles)

ΔAEX ∼ ΔADY   (AA similarity)

AE/AD = EX/DY = AX/AY

[Since, corresponding sides of two similar triangles are proportional]

EX/DY = AE/AD

EX/DY = AE/2AE (As D is the midpoint of BC)

EX/DY = ½…………..…(ii)

On Dividing eqn. (i) by eqn. (ii)

BX/EX = 4/1

BX = 4EX

BE + EX = 4EX

BE = 4EX - EX

BE = 3EX

BE /EX = 3/1

BE : EX = 3:1

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Step-by-step explanation:

GIVEN : In ∆ABC , D is the midpoint of BC, E is the midpoint of AD. BE produced meets AC at X.

To prove : BE: EX = 3 : 1

PROOF :

In Δ BCX and ΔDCY

∠CBX = ΔCBY (corresponding angles)

∠CXB = ΔCYD (corresponding angles)

ΔBCX∼ΔDCY (AA similarity)

BC/DC = BX/ DY = CX/CY

[Since, corresponding sides of two similar triangles are proportional]

BX/DY = BC/DC

BX/DY = 2DC/DC

[As D is the midpoint of BC]

BX/DY = 2/1………...(i)

In ΔAEX and ΔADY,

∠AEX = ΔADY (corresponding angles)

∠AXE = ΔAYD (corresponding angles)

ΔAEX ∼ ΔADY (AA similarity)

AE/AD = EX/DY = AX/AY

[Since, corresponding sides of two similar triangles are proportional]

EX/DY = AE/AD

EX/DY = AE/2AE (As D is the midpoint of BC)

EX/DY = ½…………..…(ii)

On Dividing eqn. (i) by eqn. (ii)

BX/EX = 4/1

BX = 4EX

BE + EX = 4EX

BE = 4EX - EX

BE = 3EX

BE /EX = 3/1

BE : EX = 3:1