Question

An isosceles triangle ABC is taken in which AB =BC . BC is produced to D . D is joined to A . AC and CD are equal . Prove that angle BAD : angle ADB = 3:1​

Answer 1

Answer:

In the diagram, we have two Isosceles triangles. For ΔABC, AB= BC and for ΔACD, AC= CD

In isosceles triangle, the two angles opposite to the equal sides are also equal. So, for ΔABC, ∠BAC = ∠ACB and for ΔACD, ∠CAD = ∠ADC

As ∠ACB is outside angle of ΔACD ,

so ∠ACB = ∠CAD + ∠ADC

⇒ ∠ACB = 2× ∠ADC (As, ∠CAD = ∠ADC )

⇒ ∠BAC = 2× ∠ADC (As, ∠BAC = ∠ACB )

Now, according to the diagram,

∠BAD - ∠CAD = ∠BAC

⇒ ∠BAD - ∠ADC = 2× ∠ADC [As, ∠CAD = ∠ADC and ∠BAC = 2× ∠ADC]

⇒ ∠BAD = 3× ∠ADC

⇒ ∠BAD = 3× ∠ADB [As, ∠ADC and ∠ADB are same angles]

⇒  ∠BAD/∠ADB=3:1

      (Proved)

Answer 2

Step-by-step explanation:

Here triangle ABC is taken

AB = BC

So by making a pt.D we get a quadrilateral

As AB : BC = 1:1 as they are similiar

So, angle D is half of the angle B which makes the total of ratio 3:1