Question

In the given figure OA is equal to OC, AB is equal to BC, Prove that Angle AOB is equal 90,
Triangle AOD is congruent to Triangle COD,
AD = CD​

Answer 1

Answer:

Step-by-step explanation:

GIVEN: OA=OC AND AB=BC

(I) TO PROVE: ∠AOB=90°

TAKE ΔAOB AND ΔBOC

OA=OC (GIVEN)

AB=BC(GIVEN)

OB=OB(COMMON)

SO, ΔAOB≅ΔBOC BY SSS RULE

THEN ∠AOB =∠BOC ---->(1)

WHICH MEANS ∠AOD = ∠COD-------->(2)

AOC IS A LINE AND OB IS A LINE MEETING IT

SO, ∠AOB+∠BOC=180°

FROM (1)

∠AOB+∠AOB=180°

2(∠AOB)=180°

∠AOB=(180/2)°

∠AOB = 90°

HENCE PROVED

(II) TO PROVE: ΔAOD≅ΔCOD

FROM (2)

∠AOD=∠COD

SO AD=CD(SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL)

OD IS COMMON LINE

SO ΔAOD=ΔCOD

HENCE PROVED

(III) TO PROVE: AD=CD

AD=CD(SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL)