Question

hi friends
I challenge u ....Can you solve this problem.....♡♡​

Answer 1

SolutioN :

( i ) D is the mid - point of AC.

• M is the mid - point of AB. ( Given )
• MD || BC

✡ D is the mid - Point of AC ( Converse of mid point Theorem )

$\rule{200}3$

( ii ) MD ⊥ AC.

• We have give ( DM || BC ) and AC is the transversal.

→ ∠MDC + ∠DCB = 180°.

→ ∠MDC + 90° = 180°.

→ ∠MDC = 180° - 90°.

→ ∠MDC = 90°.

✎ So, We can say that

✡ MD ⊥ AC.

$\rule{200}3$

( iii ) CM = MA = 1/2 AB.

Construction : Join MC.

☛ In ∆AMD and ∆CMD.

→ AD = CD ( Mid - Poin of CA )

→ ∠D = ∠D ( Each angle 90° )

→ MD = MD ( Common )

∆ AMD ≅ ∆ CMD.

✎ We can also say that,

AM = CM ( by Cpct)

• And, AB = AM + BM
• ( AM = BM )
• AB = AM + AM
• AB = 2AM
• AM = 1/2 AB.

✡ CM = MA = 1/2 AB.

Hence Proved.

Answer 2

Answer:

above answer is absolutely correct