Given: \overline{AC} \perp \overline{BD} AC ⊥ BD and \overline{BD} BD bisects \overline{AC}. AC . Prove: \triangle ABD \cong \triangle CBD△ABD≅△CBD.

By SAS (Side-Angle-Side) concurrency rule.

Step-by-step explanation:

Given:

and

To prove:

Solve,

In  ΔABD  and ΔCBD  

BD = BD    [BD is common side for both the triangles]

∠ADB = ∠CDB    [, therefore the value of both the angles are of 90 degrees]

AD = CD          []

Therefore,

By SAS (Side-Angle-Side) concurrency rule.

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The Proof for

△ABD ≅ △CBD is below

Step-by-step explanation:

Given:

AD = CD      .........BD bisect AC

To Prove:

△ABD ≅ △CBD

Proof:

In  ΔABD  and ΔCBD  

BD ≅ BD              ....……….{Reflexive Property}

∠ADB ≅ ∠CDB    …………..{Measure of each angle is 90°( )}

AD ≅ CD              ....……….{ }

ΔABD  ≅ ΔCBD   .......….{By Side-Angle-Side Congruence test}  ...Proved

where is the shape my dude?

step-by-step explanation:

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