Question

in the figure AB =AC is a point on AC and E on AB such that AD=ED=EC=BC. PROVE THAT
1. angleA:angleB=1:3
2.angle AED =angleBCE

Answer 1

Answer:

Step-by-step explanation:

Answer 2

1. ∠A : ∠B = 1:3
2. ∠A : ∠B = 1:3 ∠AED = ∠BCE

GIVEN

• AB =AC
• D is a point on AC and E on AB such that AD=ED=EC=BC.

TO FIND

1. ∠A:∠B=1:3
2. ∠AED =∠BCE

SOLUTION

We cab simply solve the above problem as follows;

In ΔABC

AB = AC (Given)

So,

∠ABC = ∠ACB (Angles of equal sides are equal)

In ΔADE

AD = ED (Given)

∠DAE = ∠ADE

Let ∠A = x

Therefore,

∠A = x = ∠DAE = ∠DEA (Equation 1)

∠EDA + x + x = 180°

∠EDA = 180-2x

Now,

∠EDC = 180- (180-2x) = 2x (Equation 2)

In ΔEDC

DE = EC (Given)

Therefore,

∠EDC =∠ECD = 2x

And,

∠DEC = 180 + 4x

∠CEB = 180- (x+180-4x)

= 3x (Equation 3)

In ΔECB

EC = BC (Given)

∠CEB = ∠CBE = 3x (Equation 4)

Now,

Ratio of ∠A and ∠B = x : 3x = 1:3

Hence, Proved.

Since,

AB = AC

So,

∠ABC = ∠ACB

Putting the values of ∠ABC and ∠ACB

3x = ∠ECB + 2x

Bringing the like terms together;

3x-2x = ∠ECB

x = ∠ECB

From equation 1

∠DEA = ∠ECB

Hence, Proved

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