Question

plz do this one hurry 13 number ​

Answer 1

Answer:

${\bold{\red{\huge{Your\:Solution}}}}$

Step-by-step explanation:

Δ  $ABC\:is\:a\:right\:angled\:triangle.$

∠ABC = 90°.

$A\:circle\:is\:drawn\:with\:AB\:as\:diameter\:intersecting\:AC\:in\:P,\\ PQ\:is\:the\:tangent\:to\:the\:circle\:which\:intersects\:BC\:at\:Q.$

$Join\:BP.$

$PQ\:and\:BQ\:are\:tangents\:drawn\:from\:an\:external\:point\:Q.$

∴ PQ = BQ   -------------- (1)  (Length of tangents drawn from an external point to the circle are equal)

⇒ ∠PBQ = ∠BPQ    (In a triangle, angles opposite to equal sides are equal)

$Given\:that,\:AB\:is\:the\:diameter\:of\:the\:circle.$

∴ ∠APB = 90°  (Angle in a semi-circle is a right angle)

∠APB + ∠BPC = 180°   (Linear pair)

∴ ∠BPC = 180° – ∠APB = 180° – 90° = 90°

$Consider$   ΔBPC,

∠BPC + ∠PBC + ∠PCB = 180°  (Angle sum property of a triangle)

∴ ∠PBC + ∠PCB = 180° – ∠BPC = 180° – 90° = 90°  ----------- (2)

∠BPC = 90°

∴ ∠BPQ + ∠CPQ = 90°    ...(3)

$From\:equations\:(2)\:and\:(3),\:we\:get$

∠PBC + ∠PCB = ∠BPQ + ∠CPQ

⇒ ∠PCQ = ∠CPQ  (Since, ∠BPQ = ∠PBQ)

$Consider$   ΔPQC,

∠PCQ = ∠CPQ

∴  $PQ = QC ----------- (4)$

$From\:equations\:(1)\:and\:(4),\:we\:get$

$BQ = QC$

$Therefore,\:tangent\:at\:bisects\:the\:side\:BC.$

Answer 2

Answer:

Step-by-step explanation:

Answer is in attachment;