Question

Help please!!
complete these proofs.

Answer 1

Explanation:

Please Identify the proofs for I'll provide the solution independently.

Given triangle: Isosceles with AC perpendicular BD

To prove: BC = CD.

Proof:

In ∆ACD And ∆ ACB,

<B = <D( Given, AB = AC and ∆ABD is Isosceles so their base angles are equal.)

<ACD = <ACB = 90°(Each of right angle, AC perpendicular to BD)

AC = AC(Common)

BY AAS rule of congruency,

∆ACD Congruent to ∆ ACB.

By Congruent parts of corresponding triangles,

BC = CD so AC bisects BD.

Answer 2

1. Given

2. Base angle of isosceles triangle

3. ∡ACB and ∡ ACD are rt. angle

4. From statement 2 or Definition of right angled triangle

5. ASA axiom

6. BC ≅ CD

7. AC bisects BD( From statement 6 and property of bisector

Explanation:

Given: Isosceles ΔABD with AB ≅ AD; AC ⊥BD

Prove: AC bisects BD.

Statements (Reason in bracket)

1. Isosceles ΔABD with AB ≅ AD; AC ⊥BD (Given)

2. ∡B ≅ ∡D (Base angle of isosceles triangle)

3. ∡ACB and ∡ ACD are rt. angle (Definition of ⊥ lines)

4. ΔACB ≅ ΔACD are rt. Δs. (From statement 2 or Definition of right angled triangle)

5. ∴ ΔACB ≅ ΔACB. (ASA axiom)

Note: Here may be RHS axiom of congruency but one side is not given, so i used ASA axiom of congruence)

6. BC ≅ CD (CPCTC)

7. ∴ AC bisects BD( From statement 6 and property of bisector)

`\textsf{Hope This Helps. Good Luck !!}`