Question

Consider triangle ABD such that angle ADB = 20° and C is a point on BD such that AB = AC and CD = CA. Then the measure of angle ABC is​

Answer 1

  Given:

       $ADB = 20^0$

      C is a point on BD such that

       $AB = AC$

       $CD = CA$

    ∴ $\angle ABC = 40^0$

Answer 2

Answer:

The measure of angle  ∠ABC is  40°

Step-by-step explanation:

Explanation:

Given ,  ABD is a triangle .

∠ ADB = 20

and C  is the on BD and

AB = AC  and CD = CA .

Step 1:

Now , in Δ ACD

AC = CD           (Given in the question)

∴ ∠DAC = ∠ADC = 20 °         (∠ADC = 20 ° given )

We know that sum of angle of triangle is equal to 180°

⇒ ∠ADC +∠ACD +∠ DAC = 180 °

⇒20 + ∠ACD + 20 = 180 °

⇒∠ACD = 180 - 40 = 140 °

Step 2:

Now , ∠ACD + ∠ACB = 180°    (Linear Pair )

⇒ 140 +∠ ACB = 180°    (∠ACD = 140 from step 1)

⇒∠ACB = 180 - 140 = 40 °

But given that , AB = AC

Therefore , ∠ACB = ∠ABC = 40°

Final answer :

Hence ,  ∠ABC  is equal to 40 °

#SPJ3