Question

In the given figure,
AD=DB = DE
angle EAC=angle FAC and
angle F=90°
Prove that angleAEB=90°
∆ACG is isosceles. ​

Answer 1

Answer:

Angle aeb is equal to 90 bcs.

Step-by-step explanation:

Ed is prependicular on line ab

Answer 2

Answer:

Correct option is

A

True

In △DBE,

DB=BE

Hence, ∠DBE=∠DEB=x ....(I)

In △DAE,

DA=AE

Hence, ∠DAE=∠DEA=y ...(II)

Now, In △ABE,

∠ABE+∠BAE+∠AEB=180

∘

(Sum of angles of triangle)

∠ABE+∠BAE+∠BED+∠DEA=180

∘

x+x+y+y=180

∘

x+y=90

∘

∠DEB+∠DEA=90

∘

∠AEB=90

∘

Hence, ∠AEB=∠AEC=90

∘

In △AEF,

Sum of angles=180

∘

∠AEF+∠EAF+∠EFA=180

∘

∠AEF+∠EAF+90

∘

=180

∘

∠AEF=90

∘

−∠EAF

We know, ∠AEC=90

∘

∠AEF+∠FEC=90

∘

90

∘

−∠EAF+∠FEC=90

∘

or ∠EAF=∠FEC

Now, In △CEA,

Sum of angles =180

∘

∠AEC+∠EAC+∠ACE=180

∘

∠ACE=90

∘

−∠EAC ..(III)

Now, in △GCE

Sum of angles=180

∘

∠GCE+∠GEC+∠CGE=180

∘

(90

∘

−∠EAC)+2∠EAC+∠CGE=180

∘

∠GCE=90

∘

−∠EAC ...(IV)

hence, from III and IV,

∠GCE=∠ACE

or GCE is an Isosceles triangle