Prove that the ratio of the areas of two similar triangle is equal to the ratio of the squares of their corresponding sides.
OR

Prove the length of tangents drawn from an external point to a circle are equal.

Question

Philoid · Accepted Answer

Let ΔPQR and ΔABC be two similar triangles,

[Corresponding sides of similar triangles are in the same ratio] [1]

And as corresponding angles of similar triangles are equal

∠A = ∠P

∠B = ∠Q

∠C = ∠R

Construction: Draw PM ⏊ QR and AN ⏊ BC

In ΔPQR and ΔABC

∠PMR = ∠ANC [Both 90°]

∠R = ∠C [Shown above]

ΔPQR ~ ΔABC [By Angle-Angle Similarity]

[Corresponding sides of similar triangles are in the same ratio] [2]

Now, we know that

Area of a triangle

Therefore,

[From 2]

[From 1]

Hence, Proved.

OR

Let us consider a circle with center O. PQ and PR be two tangents from an external point P to the circle.

To Prove: PQ = PR

Construction: Join OP, OR and OQ.

Proof:

Consider ΔOPQ and ΔOPR

∠OQP = ∠ORP [Both 90°, ∵ tangent at a point on a circle is perpendicular to the radius through point of contact]

OP = OP [Common]

OQ = OR [Radii]

⇒ ΔOPQ ≅ ΔOPR [By RHS congruency criterion]

⇒ PQ = PR [CPCT]

Hence, proved!

Prove that the ratio of the areas of two similar triangle is equal to the ratio of the squares of their corresponding sides.ORProve the length of tangents drawn from an external point to a circle are equal.