Question

20. The adjoining figure shows an isosceles AOAB with sides OA = AB = 10 units<br />and OB=12 units. Find the coordinates of the vertices of triangle AOB​

Answer 1

Step-by-step explanation:

Given: A rhombus ABCD.

 

To prove: Diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

 

Proof:

In ∆ABC,

AB = BC         (Sides of rhombus are equal.)

∠4=∠2          (Angles opposite to equal sides are equal.)     ...(1)

Now,

AD∥BC          (Opposite sides of rhombus are parallel.)

AC is transversal.

So, ∠1=∠4        (Alternate interior angles)          ...(2)

From (1) and (2), we get

∠1=∠2  

Thus, AC bisects ∠A.

Similarly,

Since, AB∥DC and AC is transversal.

So, ∠2=∠3     (Alternate interior angles)     ...(3)

From (1) and (3), we get

∠4=∠3

Thus, AC bisects ∠C.

Hence, AC bisects ∠C and ∠A

In ∆DAB,

AD = AB                   (Sides of rhombus are equal.)

∠ADB=∠ABD          (Angles opposite to equal sides are equal.)     ...(4)

Now,

DC∥AB                        (Opposite sides of rhombus are parallel.)

BD is transversal.

So, ∠CDB=∠DBA        (Alternate interior angles)          ...(5)

From (4) and (5), we get

∠ADB=∠CDB  

Thus, DB bisects ∠D.

Similarly,

Since, AD∥BC and BD is transversal.

So, ∠CBD=∠ADB     (Alternate interior angles)     ...(6)

From (4) and (6), we get

∠CBD=∠ABD

Thus, BD bisects ∠B.

Hence, BD bisects

Answer 2

${\Huge{ \red{\underline{\underline{\bf{\maltese \pink{ Ànswer: \: }}}}}}}$

REFER the ATTACHMENTS