Question

ABCD is a parallelogram. The bisectors of a consecutive angles A and B intersect at O. Show that ∠AOB=90°

Answer 1

$\huge\mathcal\pink{ANSWER:-}$

Use Properties of Triangles

Step-by-step explanation:

Given, ABCD is a Parailelogram

where AD is parallel to BC and AB is the transversal. Now, As ABAD & ACBA = 180 (Angles on the same side of transversal) As per the question,

i/BAD + iZCBA = 3(180)

= 41+ 22=90

So, In AAOB,

By Angle Sum Property of Triangles,

Z1+ 22+ ZAOB = 180

= ZAOB = 180 41 22

So, ZAOB = 90

Hence, Bisectors of any two consecutive angles of a parallelogram intersect at Right Angles .

Answer 2

Answer:

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Step-by-step explanation:

Use Properties of Triangles

Given, ABCD is a Parallelogram

where AD is parallel to BC and AB is the transversal.

Now, As ΔBAD & ΔCBA = 180 (Angles on the same side of Transversal)

As per the question,

$\frac{1}{2} ∠BAD + \frac{1}{2} ∠CBA = \frac{1}{2} (180) \\⇒∠1 + ∠2 = 90°$

So, In Δ AOB

By Angle Sum Property of Triangles,

$∠1 + ∠2 + ∠AOB = 180° \\ ⇒∠AOB = 180⁰ - ∠1 - ∠2 \\ So, ∠AOB = 90°$

Hence, Bisectors of any two consecutive angles of a parallelogram intersect at Right Angles