Mathematics — Standard (041) — CBSE Class 10 board question

Given: △ABC with ∠B = 90°. A circle with AB as diameter intersects AC at P. Let the tangent at P meet BC (extended if needed) at point Q.

To Prove: Q is the midpoint of BC, i.e., QB = QC.

Proof:

Since AB is the diameter, ∠APB = 90° (angle in a semicircle).

∴ BP ⊥ AC, which means BP is tangent to the circle? — No. Since ∠APB = 90°, and ∠ABC = 90° (given), BP ⊥ AC.

Now, the tangent at P and the chord PA make an angle equal to the angle in the alternate segment:

∠QPА = ∠PBA (tangent-chord angle = angle in alternate segment) … (1)

Since QP is tangent and QB is a secant from external point Q:

QB² = QP² — actually, use tangent lengths from Q.

From Q, QP is tangent and QB, QC involve the triangle. Let's use the tangent-from-external-point property:

Since ∠APB = 90°, PB ⊥ AC. Also QP = QB (tangent-chord angle gives △QPB isosceles):

∠QPB = ∠PBA (alternate segment theorem)
∠QPB = ∠QBP (since ∠PBA = ∠QBP, same angle)

∴ △QPB is isosceles ⟹ QP = QB … (i)

Also, QP is tangent from Q, and QC passes through the circle — but using the property: QP² = QB · QC (power of a point) … (ii)

From (i): QP = QB, substituting in (ii):
QB² = QB · QC ⟹ QB = QC

∴ The tangent at P bisects BC. Hence proved.

Source: Chapter 10, Sections 10.2–10.3

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• The key steps examiners look for: (1) ∠APB = 90° (angle in semicircle), (2) applying the tangent-chord angle (alternate segment theorem) to show △QPB is isosceles giving QP = QB, (3) applying power of a point (QP² = QB·QC) to conclude QB = QC.
• State the alternate segment theorem clearly — it is the core theorem being applied here.
• Many students forget to establish QP = QB first before using power of a point. Both steps together complete the proof.
• The answer above is approximately the right length and logical depth for a 4-mark proof question.