Exercise 3 – Animated Solutions മൂന്നാംവഴി
Four questions that use Key idea 5 (two sides in the same ratio + equal included angle ⇒ similar), together with Pythagoras. At the end we settle the research task about heights, medians and bisectors. Attempt each question first, then reveal and play.
- Key idea 5 (SAS): if two sides of a triangle are k times two sides of another and the angles between them are equal, then the third sides are in ratio k and the other angles are equal.
- Key idea 3 (SSS): all three sides in ratio k ⇒ same angles.
- Pythagoras: in a right triangle, (leg)2 + (leg)2 = (hypotenuse)2.
Question 1 · Right triangles: legs in ratio ⇒ hypotenuses in ratio
Question: Prove that if the perpendicular sides (legs) of two right triangles are scaled by the same factor, their hypotenuses are scaled by the same factor.
Let the legs be a, b and ka, kb. The angle between the legs is 90° in both triangles. Two sides in ratio k with equal included angle ⇒ Key idea 5 ⇒ the third sides (the hypotenuses) are in ratio k. Check with Pythagoras: √((ka)² + (kb)²) = √(k²(a² + b²)) = k√(a² + b²).
Question 2 · Any two sides in ratio ⇒ the third too
Question: Prove that if any two sides of two right triangles are in the same ratio, the third sides are also in that ratio.
Case 1 – the two legs: that is Question 1.
Case 2 – one leg and the hypotenuse: legs a and hypotenuse c in one triangle; ka and kc in the other. The other legs are √(c² − a²) and √((kc)² − (ka)²) = √(k²(c² − a²)) = k√(c² − a²). So the third sides are in ratio k too.
Now all three pairs are in ratio k ⇒ Key idea 3 ⇒ the triangles have the same angles: they are similar. For right triangles, two pairs of sides are enough.
Question 3 · Enlarging from a point inside
Question: Draw a triangle and mark a point inside it. Join the vertices to this point, and extend each of these lines beyond the vertex by half its length. Join the ends. Prove that the sides of the big triangle are 1½ times the sides of the original.
Call the point P and the new points A′, B′, C′. Then PA′ = PA + ½PA = 1½ PA, and similarly PB′ = 1½ PB, PC′ = 1½ PC.
In triangles PAB and PA′B′ the angle at P is common, and PA′ = 1½ PA, PB′ = 1½ PB. By Key idea 5, A′B′ = 1½ AB (and A′B′ ∥ AB). Exactly the same argument with triangles PBC, PB′C′ and PCA, PC′A′ gives B′C′ = 1½ BC and C′A′ = 1½ CA.
Orange dots can be dragged: A, B, C and the inside point P. The outer triangle is always built with PA′ = 1½ PA, PB′ = 1½ PB, PC′ = 1½ PC. Watch all three side ratios stay at 1.5 and the angles stay equal.
This "rays from a point" method is how a pantograph enlarges a drawing, how GeoGebra's Dilate from Point tool works, and why a balloon printed with a picture keeps the picture's shape as it inflates. Unlike the circle method of Section 3, it works for any shape – see Question 4.
Question 4 · The same trick for a quadrilateral
Question: The lines joining a point inside a quadrilateral to its vertices are extended outward by the same scale; joining the ends gives another quadrilateral.
(i) Prove that the sides of the big quadrilateral are the sides of the small one scaled by the same factor.
(ii) Prove that the two quadrilaterals have the same angles.
(i) Let the point be P, the scale k, and the new vertices A′, B′, C′, D′ with PA′ = k·PA etc. Triangles PAB and PA′B′ share the angle at P and have two sides in ratio k ⇒ Key idea 5 ⇒ A′B′ = k·AB. The same for PBC/PB′C′, PCD/PC′D′ and PDA/PD′A′. So all four sides are scaled by k.
(ii) Key idea 5 also says the other angles of the similar triangles are equal: ∠PA′B′ = ∠PAB and ∠PA′D′ = ∠PAD. Adding, ∠D′A′B′ = ∠D′A′P + ∠PA′B′ = ∠DAP + ∠PAB = ∠DAB. The same at B, C and D. So the quadrilaterals have the same angles.
There, scaling only the four sides of a quadrilateral did not fix its angles. Here the angles are fixed – because the point method scales the whole figure at once (every diagonal too), not just the sides.
Research task · Heights, medians and bisectors ഗവേഷണം
Question (page 118): Do the heights, medians and angle bisectors of similar triangles change by the same scale factor as the sides?
Yes, all three. Take two similar triangles with scale factor k (all sides × k, same angles).
- Height AD ⊥ BC and A′D′ ⊥ B′C′: triangles ABD and A′B′D′ have a right angle and the equal angles ∠B = ∠B′ ⇒ same angles ⇒ A′D′/AD = A′B′/AB = k.
- Median to the midpoint M of BC: B′M′ = ½B′C′ = k·½BC = k·BM, and A′B′ = k·AB with the equal included angle ∠B = ∠B′ ⇒ Key idea 5 ⇒ A′M′ = k·AM.
- Bisector AD of ∠A: ∠BAD = ½∠A = ½∠A′ = ∠B′A′D′ and ∠B = ∠B′ ⇒ triangles ABD, A′B′D′ have the same angles ⇒ A′D′ = k·AD.
In fact every matching length in similar figures scales by k – which is why the area scales by k².
Check yourself
- Right triangles: two pairs of sides in the same ratio are enough (Pythagoras supplies the third).
- Enlarging from a point: PA′ = k·PA for every vertex ⇒ every side × k, every angle unchanged – for triangles and any polygon.
- Heights, medians and angle bisectors of similar triangles scale by the same factor as the sides.
Next: Real-world applications – measure a tree with its shadow, and more ▶