In the given figure, $A B = B C$ and $A C = C D$ . Prove that: $∠ B A D : ∠ A D B = 3 : 1$ .

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Solution

Let $∠ A D B = x$

In $△ A C D, A C = C D$

$\Rightarrow ∠ C A D = ∠ C D A = x$

and, $∠ A C B = ∠ C A D + ∠ C D A = x + x = 2 x$

$\Rightarrow ∠ B A C = ∠ A C B = 2 x . (∵$ In $A B C, A B = B C)$

$∴ ∠ B A D = ∠ B A C + ∠ C A D$

$= 2 x + x = 3 x$

And, $\frac{∠ B A D}{∠ A D B} = \frac{3 x}{x} = \frac{3}{1}$

i.e., $∠ B A D : ∠ A D B = 3 : 1$ $[h e n c e p r o v e d]$

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In the given figure, AB=BC and AC=CD. Prove that: ∠BAD:∠ADB=3:1.

Let ∠ADB=xIn △ACD,AC=CD⇒∠CAD=∠CDA=xand, ∠ACB=∠CAD+∠CDA=x+x=2x⇒∠BAC=∠ACB=2x.(∵ In ABC,AB=BC)∴∠BAD=∠BAC+∠CAD=2x+x=3xAnd, ∠BAD∠ADB=3xx=31i.e., ∠BAD:∠ADB=3:1 [henceproved]

In the given figure, $A B = B C$ and $A C = C D$ . Prove that: $∠ B A D : ∠ A D B = 3 : 1$ .