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Question Bank - P2 — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsQuestion Bank - P24 Marks
Question
In the figure ∆ABC is an equilateral triangle.The angle bisector of ∠? will intersect the circumcircle ∆ABC at point P. Then prove that : CQ = CA.
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Answer

$\triangle ABC$ is an equilateral triangle.
$\therefore \angle ABC =\angle ACB =\angle BAC =60^{\circ} ....(i) [Angles of an equilateral triangle]$
$\angle CBP =\frac{1}{2} \angle ABC$
$\therefore \angle CBP =\frac{1}{2} \times 60^{\circ}$
$\therefore \angle CBP =30^{\circ}$
$\angle C B P=\angle C A P=30^{\circ} ....[Angles inscribed in the same arc]$
$\therefore \angle CAQ =30^{\circ}$
In $\triangle ABQ$,
$\angle B A Q=\angle B A C+\angle C A Q ....[Angle addition property]$
$\therefore \angle BAQ =60^{\circ}+30^{\circ}$
In $\triangle A B Q$,
$ \angle BAQ =\angle BAC +\angle CAQ \quad \ldots . .[\text { Angle addition property }]$
$\therefore \angle BAQ =60^{\circ}+30^{\circ} \quad \ldots . .[\text { From (i) and (ii) }]$
$\therefore \angle BAQ =90^{\circ} $
Also, $\angle A B Q=60^{\circ}$ [From (i) and B-C-Q]
$\therefore \angle BQA =30^{\circ}$
$\therefore \angle C Q A=30^{\circ}$
In $\triangle CQA$,
$\angle C A Q=\angle C Q A [From (ii) and (iii)]$
$\therefore CQ = CA [Converse of isosceles triangle theorem]$

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