If Triangle Abc is isosceles with …
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Sia ? 6 years, 4 months ago
Draw AD {tex}\perp{/tex} BC
In {tex} ADB \ and \ ADC {/tex}
{tex}AB = AC{/tex} [Given]
{tex}AD = AD{/tex}
{tex}ADB = ADC [Each \ 90^o]{/tex}
{tex}\therefore{/tex} {tex}\triangle \mathrm { ADB } \cong \triangle \mathrm { ADC }{/tex}
{tex}\Rightarrow{/tex} {tex}BD = CD{/tex}
{tex}\therefore{/tex} AC passes through O,centre of the circle
{tex}\therefore{/tex} Perpendicular bisector of the chord passes through the centre of the circle
Now OA {tex}\perp{/tex} PQ (radius through the point of contact)
{tex}\therefore{/tex} {tex}\angle{/tex}PAO = 90o
Also {tex} ADB = 90^o{/tex}
{tex}\therefore{/tex} {tex}\angle \mathrm { PAO } + \angle \mathrm { ADB } = 180 ^ { \circ }{/tex}
{tex}\therefore{/tex} AP||BC
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