Question

solve the second one as soon as possible​

Answer 1

${\large\orange{\mathbb\fcolorbox{pink}{maroon}{★Solution ★:-}}}$

$\sf{\bold{\green{\underline{\underline{Given}}}}}:$

• PA is the Tangent.
• AB is a Chord.

$\sf {\bold{\green{\underline{\underline{To \: prove :-}}}}}$

• ∆ PAD is an isosceles Triangle
• $∠CAD = \frac{1}{2} (∠ \: PBA - \: ∠PAB)$

$\sf {\bold{\green{\underline{\underline{Required \: Solution :- }}}}}$

So, ∠ PAB = ∠ C -----(1)

[Angle are alternate Segment ]

AD is the Bisector of ∠BAC

$∠1 = ∠2 \: \: - - (2)$

In, ∆ADC ,

Exterior ∠AOP = $∠C + ∠1$

Exterior ∠ADP = $∠PAB+ ∠C \\ \implies∠PAD$

Therefore,

☆ ∆ PAD is an Isosceles triangle.

Now,

In a ∆ABC

Exterior ∠ PBA = $∠C + ∠BAC$

∠BAC = ∠PBA - ∠C

We can write it like this ,

$\boxed{∠1 +∠2 = ∠PBA - ∠ PAB} \:$

$\sf\colorbox{green}{ From Equation 1st }$

$\implies2∠ 1 = ∠ PAB -∠ PAB \\ \implies∠1 = \frac{1}{2} (∠ PAB -∠ PAB) \\ : \implies\boxed{∠CAD = \frac{1}{2} (∠ PAB -∠ PAB)}$

☆ Hence Proved!!

_____________________________

$\sf\blue{hope \: this \: helps \: you!! \: }$

@Mahor2111

_________♬♩♪♩ ♩♪♩♬__________

Answer 2

⟹2∠1=∠PAB−∠PAB

⟹∠1= 21 (∠PAB−∠PAB)

:⟹ ∠CAD= 21 (∠PAB−∠PAB)