Question

(viii) A perpendicular AD is drawn on the chord BC of the circle at the point A on its circumference. If AE is the diameter of the circle, prove that ΔBAD~=ΔEAC.
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Answer 1

$\Large{\underline{\underline{\bf{Solution :}}}}$

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★ Given :

A perpendicular AD is drawn on the chord BC of the circle at the point A on its circumference. If AE is the diameter of the circle, prove that ΔBAD~=ΔACD.

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★ To Prove :

We have to prove that ∆BAD $\cong$ ∆ACD

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★ Proof :

In ∆ BAD and ∆ ACD

$\sf{\angle BAD = \angle DAC \: \: \: \: \: \: \: (\because \: perpendicular)} \\ \\ \sf{BD = CD \: \: \: \: \: \: \: (Tangent)} \\ \\ \sf{AD = AD \: \: \: \: \: \: \: (Common)} \\ \\ \sf{\therefore \: \triangle BAD \: \cong \: \triangle ACD}$

Hence proved

Answer 2

Answer:

$\huge\underline{ \underline{ \red{your \: \: answer}}}$

Step-by-step explanation:

$\sf\red{A \: perpendicular \: AD \: is \: drawn \: on \: the \: chord \: BC} \\ \sf\red{of \: the \: \: circle \: at \: the \: point \: A \: on \: its \: circumference. } \\ \sf\red{ If \: AE \: is \: the \: \: diameter \: of \: thecircle, \: prove \: that \: } \\ \sf\red{.} \\ \\ \sf\blue{we \: have \: to \: prove \: that \: ΔBAD \congΔACD} \\ \\ \sf\pink{} \\ \\ \sf\pink{now \: ΔBAD \: and \: ΔACD} \\ \\ \sf\green{ \angle \: BAD = \angle DCA( \because \: parpenicular)} \\ \\ \sf\orange{AD = AD(common)} \\ \\ \sf\blue{BD =CD(tangent)} \\ \\ \sf\pink{ \underline{ \therefore \: ΔBAD \congΔACD}} \\ \\ \\ \\ \sf\purple{ \underline{proved}}$

$\large\boxed{ \fcolorbox{green}{yellow}{hope \: it \: help \: you}}$