Question

Q9. ln triangle ABD, C is any point on BD such that AB = BC and AC =CD. Prove that Angle BAD : Angle ADB = 3 : 1.

Answer 1

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]

⇒ $\frac{\angle BAD}{\angle ADB}= \frac{3}{1}$

So, $\angle BAD : \angle ADB= 3:1$

Answer 2

Answer:

Step-by-step explanation: ABCD IS CYCLIC JOIN BD AND AC. INCENTRE , I LIES ON BD PROVE AB= BC=CI