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Let AB be a line segment, P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B. PQ intersect AB at a point C. Then,
Option: 1
$\angle A C P=90^{\circ}$

Option: 2
$\angle A P Q=\angle B P Q$

Option: 3
$\angle A P C=\angle B P C$

Option: 4
All of the above

Answers (1)

best_answer

$\begin{align*} &\text{In } \triangle PAQ \text{ and } \triangle PBQ, \text{ we have:} \\ &PA = PB \quad \text{(given)} \\ &QA = QB \quad \text{(given)} \\ &PQ = PQ \quad \text{(common)} \\ \therefore &\triangle PAQ \cong \triangle PBQ \quad \text{(by SSS-criteria)} \\ \therefore &\angle APQ = \angle BPQ \qquad \ldots (i) \end{align*}$

$\\\text{Now, in } \triangle P A C\text{ and } \triangle P B C,\text{ we have} \\ P A=P B \quad \text{ (given)} \\\angle A P C=\angle B P C \qquad {[\because \angle A P Q=\angle B P Q \text { in }( i )]} \\ P C=P C \text{( common )} \\ \\\therefore \quad \triangle P A C \cong \triangle P B C \quad \text { (by SAS-criteria) }$

$\\\therefore \quad A C=B C\qquad \text{(c.p.c.t.)} \\ \text{And, } \angle A C P=\angle B C P\qquad \ldots(iii) \\\text{But, } \angle A C P+\angle B C P=180^{\circ}\text{ (linear pair) }\\\therefore \quad 2 \angle A C P=180^{\circ} \quad[\text { using }( iii )] \\\\\text{So, }\angle A C P=90^{\circ}$

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