Question

In a quadrilateral ABCD, AB = AD and BC = CD.

Prove that :

1. AC bisects ∠BAD 
2. AC is the perpendicular bisector of BD.

Answer 1

Given:

Quadrilateral ABCD with AB = AD and BC = CD.

To Prove:

1. AC bisects ∠BAD. 
2. AC is the perpendicular bisector of BD.

Proof (Step-wise):

1. Consider triangles ABC and ADC.

AB = AD   ...(Given)

BC = CD   ...(Given)

AC = AC   ...(Common side)

Thus, triangles ABC and ADC are congruent by SSS.

2. From the congruence in step 1,

Corresponding angles at A are equal:

∠BAC = ∠DAC

Therefore, AC bisects ∠BAD. 

This proves i.

3. Let O be the intersection point of AC and BD (O = AC ∩ BD).

From step 2, we have ∠BAO = ∠OAD.

Also AB = AD (Given) and AO = AO (Common).

Hence, triangles AOB and AOD are congruent by SAS.

Therefore, OB = OD.

So, AC meets BD at O and O is the midpoint of BD, AC bisects BD.

4. From the congruence of triangles AOB and AOD, we also get ∠AOB = ∠DOA.

Note that OA and OC are opposite rays (A,O,C collinear).

So, ∠AOB + ∠BOC = 180°.   ...(Linear pair)

Similarly, OB and OD are opposite rays (B,O,D collinear).

So, ∠DOA + ∠AOB = 180°.   ...(Linear pair)

Equating the two expressions and using ∠DOA = ∠AOB gives ∠AOB = ∠BOC.

Since ∠AOB + ∠BOC = 180°, each equals 90°.

Hence, ∠AOB = 90°, so AC ⟂ BD.

5. Combining step 3 (OB = OD) and step 4 (AC ⟂ BD) shows that AC passes through the midpoint of BD and meets BD at right angles.

Thus, AC is the perpendicular bisector of BD.

This proves ii.