Question

Tell me fastely guyes​

Answer 1

Answer:

Answer is in attachment

Answer 2

Given:

Right Angle Triangle ABC

AM = BM [M is the mid point of hypotenuse AB]

DM = CM

To Prove:

1. Δ AMC ≅ Δ BMD

2. ∠ DBC is a right angle [∠DBC = 90°]

3. Δ DBC ≅ Δ ACB

4. CM = $\frac{1}{2}$ AB

Proof:

In Δs AMC and BMD:

∠AMC = ∠BMD    [Vertically Opposite Angles]

AM = BM     [M is the mid-point  of Hypotenuse AB]

DM = CM     [Given]

∴ ΔAMC ≅ ΔBMD                    [S.A.S. Congruency Criteria]

=> ∠CAM = ∠DBM     [Corresponding parts of Congruent Δs]

and DB = AC      [same reason]       ----------- Equation 1

∠CAM = ∠DBM      [proven above]

But these are alternate interior angles.

=> BD ║ AC

As BD ║AC,

∠DBC + ∠ACB = 180°     [Sum of Consecutive Interior Angles is always 180°]

But ABC is a Right Angled Triangle, where ∠C = 90°

=> ∠DBC + 90° = 180°

=> ∠DBC = 90°

Hence ∠DBC is a right angle.

In Δs DBC and ACB:

DB = AC     [From Equation 1]

BC = BC     [Common]

∠DBC = ∠ACB     [90° each]

∴ ΔDBC ≅ ΔACB                    [S.A.S. Congruency Criteria]

=> DC = AB          [Corresponding parts of Congruent Δs]

=> DM + CM = AB

=> CM + CM = AB         [CM = DM <= given]

=> CM =  $\frac{1}{2}$ AB

Hence Proved

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