Question

No.7 . The best answer will be brainliest marked.

Answer 1

Hello buddy!

Given: ABCD is a rhombus, AC is a diagonal

To prove: AC bisects angle BCD

Proof:
AB ≅ CD (opposite sides of a parallelogram are congruent)


BC ≅ AD (opposite sides of a parallelogram are congruent)


AC ≅ AC (a line is congruent to itself.)


ΔABC ≅ ΔADC (SSS congruency)


∠1 ≅ ∠4 (corresponding parts of congruent triangles.)


∠3 ≅ ∠2 (corresponding parts of congruent triangles.)


AB ≅ BC (definition of rhombus, all sides congruent.)


ΔABC is isosceles (two sides are congruent)


∠1 ≅ ∠3 (base angles of an isosceles triangle)


∠1 ≅ ∠2 (things congruent to the same thing to each other, AC bisects ∠ABD)


∠3 ≅ ∠4 (things congruent to the same thing to each other)


AC bisects ∠BCD ( from the above cases)

Thank you
Hope it helped!

Answer 2

Hello there! Please see the picture which I have attached to this answer.

Summary:

1. Divide the Rhombus into 4 triangles
2. Prove all the 4 pairs of adjacent triangles to be congruent to each other.
3. Using CPCT, you can state that the vertex angles are bisected.

See, that's it! It is a very easy problem, and I believe you can solve it!!!

Please mark this as Brainliest if you find this answer useful!