Question

$\huge \bf Heya \: Guys$
In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Prove that the tangent to the circle at P bisects BC.
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Answer 1

Given:

• In a right triangle ABC in which ∠B = 90°.
• a circle is drawn with AB as diameter intersecting the hypotenuse AC at P.

To Prove:

• ➲ that the tangent to the circle at P bisects BC.

Solution:

★$\sf\underline{According \: To \: The\:Question:}$

$\sf\triangle \: abc \: is \: right \: angled \\ \\ \sf \angle \: abc = 90 \degree \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now,

➸PQ is the tangent which intersects BC at Q

➸PQ and BQ are tangents drawn from external point Q

➷Therefore, PQ = BQ

_____________________

$: \implies \sf \angle \: pbq = \angle \: bpq$

(Angles opposite to equal sidea are equal)

➸As, AB is the diameter of the circle

$: \implies \sf\angle \: apb \: = 90 \degree \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ : \implies \sf\angle \: apb + \angle \: bpc = 180\degree \red{(linear \: pair)} \\ \\ \\ : \implies \sf \angle \: bpc = 180 - 90 \degree \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ : \implies \blue\angle \sf \: \blue{bpc = 90\degree} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

_____________________

Now,

due to angle sum property,

$: \implies \sf \: \angle \: apb + \angle \: bpc + \angle \: pcb = 180\degree$

so,

$: \implies \sf\angle \: pcb + \angle \: pbc = 90\degree\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

_____________________

$: \implies \sf\angle \: bpc = 90\degree \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf\therefore \angle \: bpq + cpq = 90\degree \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now, these equations say that,

$: \implies \sf\angle \: pbc + pcb = \angle \: bpq + cpq \\ \\ \\ : \implies \sf\angle \: pcq = cpq\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \therefore \: { \boxed{\rm{ \: pq = cq}}}$

So,

$\sf {\blue {\bold{bc = qc}}}$

$\blue{ \underline{ \boxed{ \pink{ \mathfrak{hence \: proved \: tangent \: at \: p \: bisects \: the \: side \: bc}}}}}$

Answer 2

In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Prove that the tangent to the circle at P bisects BC.

$\sf{------------@BrainlyKingdom------------}$

Let O be the center of the given circle.

• Suppose the tangent at P meets BC at Q

• Then join BP

The aim is to show that BQ = QC.

The diagram for the given is as, (Refer to Attachment)

• ∠ABC is 90 since tangent at any point of the circle is perpendicular to the radius through the point of contact.

In ∆ABC, ∠1 + ∠5 = 90° (Angle sum property) And ∠3 = ∠1

• So, ∠3 + ∠5 = 90°

Also, ∠APB = 90° (Angle in semi circle)

• ∠3 + ∠4 = 90°

So, ∠3 + ∠5 = ∠3 + ∠4

⇒ ∠5 = ∠4

⇒ PQ = QC

Also QP = QB (tangents drawn from an internal point to a circle are equal)

So, QC = QB