Question

In the given figure, ab=bc and ac=cd, prove that angle bad : angle adb = 3: 1
The figure is a triangle abd in which ac is the mid point of it. Please answer this question

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]

⇒ $\frac{\angle BAD}{\angle ADB} = \frac{3}{1} \\ \\ So, \angle BAD: \angle ADB= 3:1$ (Proved)

Answer 2

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